Question 515 Marks
The vertices A, B, C of triangle ABC have respectively position vector $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ with respect to given origin O. Show that the point D where the bisector of $\angle{\text{A}}$ meets BC has position Vector $\vec{\text{d}}=\frac{\beta\vec{\text{b}}+\gamma\vec{\text{c}}}{\beta+\gamma}$, where $\beta=\big|\vec{\text{c}}-\vec{\text{a}}\big|$ and, $\gamma=\big|\vec{\text{a}}-\vec{\text{b}}\big|$.
AnswerLet the position vectors of A, B and C with respect to same origin, O be $\vec{\text{a}},\ \vec{\text{b}}\text{ and }\vec{\text{c}}$ respectively.
Let D be the point on BC where bisectors of $\angle\text{A}$ meets.
Let $\vec{\text{d}}$ be the position vector of D which divides CB internally in the ratio $\beta \text{ and } \gamma$, where
$\beta=\Big|\overrightarrow{\text{AC}}\Big|\text{ and }\gamma=\Big|\overrightarrow{\text{AB}}\Big|$
By section formula, the position vector of D is given by
$\overrightarrow{\text{OD}}=\frac{\beta\vec{\text{b}}+\gamma\vec{\text{c}}}{\beta+\gamma}$
Let $\alpha=\big|\vec{\text{b}}-\vec{\text{c}}\big|$
In center is the concurent point of angle bisectors an in center divides the line AD in the ratio $\alpha : \beta +\gamma$.
So, the position vector of in center is given as,
$\frac{\alpha\vec{\text{a}}+\Big(\frac{\beta\vec{\text{b}}+\gamma\vec{\text{c}}}{\beta+\gamma}\Big)(\beta+\gamma)}{\alpha+\beta+\gamma}=\frac{\alpha\vec{\text{a}}+\beta\vec{\text{b}}+\gamma\vec{\text{c}}}{\alpha+\beta+\gamma}$
View full question & answer→Question 525 Marks
Prove that: $\big(\vec{\text{a}}-\vec{\text{b}}\big).\big\{\big(\vec{\text{b}}-\vec{\text{c}}\big)\big\}=0$
AnswerWe have
$\big(\vec{\text{a}}-\vec{\text{b}}\big).\big\{\big(\vec{\text{b}}-\vec{\text{c}}\big)\times\big(\vec{\text{c}}-\vec{\text{a}}\big)\big\}$
$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{c}}-\vec{\text{b}}\times\vec{\text{a}}-\vec{\text{c}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}\big)$ (By distributive law)
$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{c}}-\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$\big(\therefore\vec{\text{c}}\times\vec{\text{c}}=0\big)$
$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)+\vec{\text{a}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)+\vec{\text{a}}\big(\vec{\text{c}}\times\vec{\text{a}}\big)-\vec{\text{b}}\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\vec{\text{b}}\big(\vec{\text{a}}\times\vec{\text{b}}\big)-\vec{\text{b}}\big(\vec{\text{c}}\times\vec{\text{a}}\big)$ (By distributive 1)
$=\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]+\big[\vec{\text{a}}\ \vec{\text{a}}\ \vec{\text{b}}\big]+\big[\vec{\text{a}}\ \vec{\text{c}}\ \vec{\text{a}}\big]-\big[\vec{\text{b}}\ \vec{\text{b}}\ \vec{\text{c}}\big]-\big[\vec{\text{b}}\ \vec{\text{a}}\ \vec{\text{b}}\big]-\big[\vec{\text{b}}\ \vec{\text{c}}\ \vec{\text{a}}\big]$
$=\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]+0+0-0-0-\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]$
$\big(\therefore\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]=\big[\vec{\text{b}}\ \vec{\text{c}}\ \vec{\text{a}}\big]\big)$
$=0$
View full question & answer→Question 535 Marks
If $\vec{\text{a}}\text{ and }\vec{\text{b}}$ are two non-collinear vectors having the same initial point. What are the vectors represented by $\vec{\text{a}}+\vec{\text{b}}\text{ and }\vec{\text{a}}-\vec{\text{b}}$.
AnswerHere, it is given that $\vec{\text{a}}\text{ and }\vec{\text{b}}$ are two non-collinear vectors having the same initial point.Let $\vec{\text{a}}=\overrightarrow{\text{AB}}\text{ and }\vec{\text{b}}=\overrightarrow{\text{AD}}$, So we can draw a parallelogram ABCD as above.
By the properties of parallelogram
$\overrightarrow{\text{BC}}=\vec{\text{b}}\text{ and }\overrightarrow{\text{DC}}=\vec{\text{a}}$
In $\triangle\text{ABC}$,
Using triangle law,
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}$
$\vec{\text{a}}+\vec{\text{b}}=\overrightarrow{\text{AC}}\ \dots(\text{i})$
In $\triangle\text{ABD}$
$\overrightarrow{\text{AD}}+\overrightarrow{\text{DB}}=\overrightarrow{\text{AB}}$
$\vec{\text{b}}+\overrightarrow{\text{DB}}=\vec{\text{a}}$
$\overrightarrow{\text{DB}}=\vec{\text{a}}-\vec{\text{b}}\ \dots(\text{ii})$
From (i) and (ii), we get that
$\vec{\text{a}}+\vec{\text{b}}\text{ and }\vec{\text{a}}-\vec{\text{b}}$ are diagonals of a parallelogram whose adjacent sides are $\vec{\text{a}}\text{ and }\vec{\text{b}}$
View full question & answer→Question 545 Marks
Find $\lambda$ for which the points A(3, 2, 1), B(4, $\lambda$, 5), C(4, 2, -2) and D(6, 5, -1) are coplanar.
AnswerThe points A, B, C and D will be coplanar if will be coplanar of any one of the following traces of vectors are coplanar:
$\vec{\text{AB}},\vec{\text{AC}},\vec{\text{AD}};\vec{\text{AB}},\vec{\text{BC}},\vec{\text{CD}};\vec{\text{BC}},\vec{\text{BA}},\vec{\text{BD}},$ etc.
It is given that $\vec{\text{AB}},\vec{\text{AC}},\vec{\text{AD}}$ are coplanar.
Thus, their scaler triple product $\big[\vec{\text{AB }}\vec{\text{AC }}\vec{\text{AD}}\big]$ is equal to zero.
Now,
Direction ratios of the $\vec{\text{PQ}}$ =(Direction ratios of vector Q) - (Direction ratios of the vector P)
Direction ratios of vector $\vec{\text{AB}}=(4-3),(\lambda-2),(5-1),\text{i. e. 1},\lambda, -2, 4$
Direction ratios of vector $\vec{\text{AC}}=(4-3),(2-2),(-2 -1),\text{i. e. } 3, 3, -2$
Direction ratios of vector $\vec{\text{AD}}=(6-3),(5-2),(-1-1),\text{i. e}. 3, 3, -2$
$\therefore\big[\vec{\text{AB}}\vec{\text{ AC }}\vec{\text{AD}}\big]=\begin{vmatrix}1&\lambda-2&4\\1&0&-3\\3&3&-2\end{vmatrix}$
$=1[0-(-9)]-(\lambda-2)[-2-(-9)]+4(3-0)=0$
$\Rightarrow7\lambda=35$
$\Rightarrow\lambda=5$
View full question & answer→Question 555 Marks
Show that the points $\text{A}\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big),\ \text{B}\big(\hat{\text{i}}-3\hat{\text{j}}-5\hat{\text{k}}\big),$ $\text{C}\big(3\hat{\text{i}}-4\hat{\text{j}}-4\hat{\text{k}}\big)$ are the vertices of a right angled triangle.
AnswerGiven the points $\text{A}\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big),\ \text{B}\big(\hat{\text{i}}-3\hat{\text{j}}-5\hat{\text{k}}\big)$and $\text{C}\big(3\hat{\text{i}}-4\hat{\text{j}}-4\hat{\text{k}}\big)$. Then, $\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A $=\hat{\text{i}}-3\hat{\text{j}}-5\hat{\text{k}}-\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$ $=\hat{\text{i}}-3\hat{\text{j}}-5\hat{\text{k}}-2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$ $=-\hat{\text{i}}-2\hat{\text{j}}-6\hat{\text{k}}$ $\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B $=3\hat{\text{i}}-4\hat{\text{j}}-4\hat{\text{k}}-\big(\hat{\text{i}}-3\hat{\text{j}}-5\hat{\text{k}}\big)$ $=3\hat{\text{i}}-4\hat{\text{j}}-4\hat{\text{k}}-\hat{\text{i}}+3\hat{\text{j}}+5\hat{\text{k}}$$=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\overrightarrow{\text{CA}}=$ Position vector of A - Position vector of C $=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}-\big(3\hat{\text{i}}-4\hat{\text{j}}-4\hat{\text{k}}\big)$ $=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}-3\hat{\text{i}}+4\hat{\text{j}}+4\hat{\text{k}}$ $=-\hat{\text{i}}+3\hat{\text{j}}+5\hat{\text{k}}$ Clearly, $\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=\vec0$Now, $\Big|\overrightarrow{\text{AB}}\Big|=\sqrt{(-1)^2+(-2)^2+(-6)^2}$
$=\sqrt{1+4+36}$ $=\sqrt{41}$ $\Big|\overrightarrow{\text{BC}}\Big|=\sqrt{(2)^2+(-1)^2+(1)^2}$ $=\sqrt{4+1+1}$ $=\sqrt{6}$ $\Big|\overrightarrow{\text{CA}}\Big|=\sqrt{(-1)^2+(3)^2+(5)^2}$ $=\sqrt{1+9+25}$ $=\sqrt{35}$ Clearly, $\Big|\overrightarrow{\text{AB}}\Big|^2=\Big|\overrightarrow{\text{BC}}\Big|^2+\Big|\overrightarrow{\text{CA}}\Big|^2$$\Rightarrow\text{AB}^2=\text{BC}^2+\text{CA}^2$
So, A, B, C forms a right angled triangle.
View full question & answer→Question 565 Marks
Prove that the given vectors are non-coplanar:
$3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\ 2\hat{\text{i}}-\hat{\text{j}}+7\hat{\text{k}}$ and $7\hat{\text{i}}-\hat{\text{j}}+23\hat{\text{k}}$
AnswerWe know that, Three vectors are coplanar if one of them vector can be expressed as the linear combination of the other two. Let, $\big(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)\\=\text{x}\big(2\hat{\text{i}}-\hat{\text{j}}+7\hat{\text{k}}\big)+\text{y}\big(7\hat{\text{i}}-\hat{\text{j}}+23\hat{\text{k}}\big)$ $=2\text{x}\hat{\text{i}}-\text{x}\hat{\text{j}}+7\text{x}\hat{\text{k}}+7\text{y}\hat{\text{i}}-\text{y}\hat{\text{j}}+23\text{y}\hat{\text{k}}$ $\big(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)\\=\big(2\text{x}+7\text{y}\big)\hat{\text{i}}+\big(-\text{x}-\text{y}\big)\hat{\text{j}}+\big(7\text{x}+23\text{y}\big)\hat{\text{k}}$ Equating the coefficient of LHS and RHS, 2x + 7y = 3 .....(i) -x - y = 1 .....(ii) 7x + 23y = -1 .....(iii) For solving (i) and (ii), Add (i) and 2 × (ii),
$\text{y}=\frac{5}5$ $\text{y}=1$ Put the value of y in equation (i), $2\text{x}+7\text{y}=3$ $2\text{x}+7(1)=3$ $2\text{x}+7=3$ $2\text{x}=3-7$ $2\text{x}=-4$ $\text{x}=\frac{-4}2$ $\text{x}=-2$ Put the value of x and y in equation (iii), $7\text{x}+23\text{y}=-1$ $7(2)+23(1)=-1$ $14+23=-1$ $37=-1$ $\text{LHS}\neq\text{RHS}$ The value of x and y do not satisfy the equation (iii), Hence, vectors are non-coplanar. View full question & answer→Question 575 Marks
If $\hat{\text{a}}$ and $\hat{\text{b}}$ are unit vectors inclined at an angle $\theta$, prove that$\tan\frac{\theta}{2}=\frac{\big|\hat{\text{a}}-\hat{\text{b}}\big|}{\big|\hat{\text{a}}+\hat{\text{b}}\big|}$
AnswerHere, $\hat{\text{a}}$ and $\hat{\text{b}}$ are unit vectors
$\big|\hat{\text{a}}\big|=\big|\hat{\text{b}}\big|=1$
$\frac{\big|\hat{\text{a}}-\hat{\text{b}}\big|^2}{\big|\hat{\text{a}}+\hat{\text{b}}\big|^2}=\frac{(\hat{\text{a}}-\hat{\text{b}})^2}{(\hat{\text{a}}+\hat{\text{b}})^2}$
$=\frac{(\hat{\text{a}})^2+(\hat{\text{b}})^2-2\hat{\text{a}}.\hat{\text{b}}}{(\hat{\text{a}})^2+(\hat{\text{b}})^2+2\hat{\text{a}}.\hat{\text{b}}}$
$=\frac{\big|\hat{\text{a}}\big|^2+\big|\hat{\text{b}}\big|^2-2\hat{\text{a}}.\hat{\text{b}}}{\big|\hat{\text{a}}\big|^2+\big|\hat{\text{b}}\big|^2+2\hat{\text{a}}.\hat{\text{b}}}$
$\frac{\big|\hat{\text{a}}-\hat{\text{b}}\big|^2}{\big|\hat{\text{a}}+\hat{\text{b}}\big|^2}=\frac{(1)^2+(1)^2-2\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\cos\theta}{(1)^2+(1)^2+2\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\cos\theta}$ $\big[\text{since }\vec{\text{a}}.\vec{\text{b}}=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\cos\theta\big]$
$\frac{\big|\hat{\text{a}}-\hat{\text{b}}\big|^2}{\big|\hat{\text{a}}+\hat{\text{b}}\big|^2}=\frac{1+1-2(1)(1)\cos\theta}{1+1+2(1)(1)\cos\theta}$
$=\frac{2-2\cos\theta}{2+2\cos\theta}$
$=\frac{2(1-\cos\theta)}{2(1+\cos\theta)}$
$=\frac{2\times\sin^2\frac{\theta}{2}}{2\times\cos^2\frac{\theta}{2}}$ $\Big[\text{Since}1-\cos\theta=2\sin^2\frac{\theta}{2},1+\cos\theta=2\cos^2\frac{\theta}{2}\Big]$
$\frac{\big|\hat{\text{a}}-\hat{\text{b}}\big|^2}{\big|\hat{\text{a}}+\hat{\text{b}}\big|^2}=\tan^2\frac{\theta}{2}$
$\tan\frac{\theta}{2}=\frac{\big|\hat{\text{a}}-\hat{\text{b}}\big|^2}{\big|\hat{\text{a}}+\hat{\text{b}}\big|^2}$
View full question & answer→Question 585 Marks
If $\vec{\text{a}}=2\hat{\text{i}}+5\hat{\text{j}}-7\hat{\text{k}},\vec{\text{b}}=-3\hat{\text{i}}+4\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}},$ compute $\big(\vec{\text{a}}\times\vec{\text{b}}\big)\times\vec{\text{c}}$ and $\vec{\text{a}}\times\big(\vec{\text{b}}\times\vec{\text{c}}\big)$ and verify that these are not equal.
AnswerGiven:
$\vec{\text{a}}=2\hat{\text{i}}+5\hat{\text{j}}-7\hat{\text{k}}$
$\vec{\text{b}}=-3\hat{\text{i}}+4\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
$\therefore\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&5&-7\\-3&4&1 \end{vmatrix}$
$=(5+28)\hat{\text{i}}-(2-21)\hat{\text{j}}+(8+15)\hat{\text{k}}$
$=33\hat{\text{i}}+19\hat{\text{j}}+23\hat{\text{k}}$
$\Rightarrow\big(\vec{\text{a}}\times\vec{\text{b}}\big)\times\vec{\text{c}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\33&19&23\\1&-2&-3 \end{vmatrix}$
$=(-57+46)\hat{\text{i}}-(-99-23)\hat{\text{j}}+(-66-19)\hat{\text{k}}$
$\Rightarrow\big(\vec{\text{a}}\times\vec{\text{b}}\big)\times\vec{\text{c}}=-11\hat{\text{i}}+122\hat{\text{j}}-85\hat{\text{k}}\dots(1)$
$\therefore\vec{\text{b}}\times\vec{\text{c}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\-3&4&1\\1&-2&-3 \end{vmatrix}$
$=(-12+2)\hat{\text{i}}-(9-1)\hat{\text{j}}+(6-4)\hat{\text{k}}$
$=-10\hat{\text{i}}-8\hat{\text{j}}+2\hat{\text{k}}$
$\Rightarrow\vec{\text{a}}\times\big(\vec{\text{b}}\times\vec{\text{c}}\big)=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&5&-7\\-10&-8&2 \end{vmatrix}$
$=(10-56)\hat{\text{i}}-(4-70)\hat{\text{j}}+(-16+50)\hat{\text{k}}$
$\Rightarrow\vec{\text{a}}\times\big(\vec{\text{b}}\times\vec{\text{c}}\big)=-46\hat{\text{i}}+66\hat{\text{j}}+34\hat{\text{k}}\dots(2)$
From (1) and (2), we get
$\big(\vec{\text{a}}\times\vec{\text{b}}\big)\times\vec{\text{c}}\neq\vec{\text{a}}\times\big(\vec{\text{b}}\times\vec{\text{c}}\big)$
View full question & answer→Question 595 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are non-zero, non-coplanar vectors, prove that the vector is coplanar:
$\vec{\text{a}}-2\vec{\text{b}}+3\vec{\text{c}},\ -3\vec{\text{b}}+5\vec{\text{c}}$ and $-2\vec{\text{a}}+3\vec{\text{b}}-4\vec{\text{c}}$
AnswerWe know that, Three vectors are coplanar if one of them can be expressed as the linear combination of other two. Let, $\vec{\text{a}}-2\vec{\text{b}}+3\vec{\text{c}}\\=\text{x}\big(-3\vec{\text{b}}+5\vec{\text{c}}\big)+\text{y}\big(-2\vec{\text{a}}+3\vec{\text{b}}-4\vec{\text{c}}\big)$ $\vec{\text{a}}-2\vec{\text{b}}+3\vec{\text{c}}\\=-3\vec{\text{b}}\text{x}+5\vec{\text{c}}\text{x}+2\vec{\text{a}}\text{y}+3\vec{\text{b}}\text{y}-4\vec{\text{c}}\text{y}$ $\vec{\text{a}}-2\vec{\text{b}}+3\vec{\text{c}}\\=\big(-2\text{y}\big)\vec{\text{a}}+\big(-3\text{x}+3\text{y}\big)\vec{\text{b}}+\big(5\text{x}-4\text{y}\big)\vec{\text{c}}$ Comparing the LHS and RHS, -2y = 1 .....(i) -3x + 3y = -2 .....(ii) 5x - 4y = 3 .....(iii) For solving (i) and $\text{y}=-\frac{1}2$ Put value of y in equation (ii), $-3\text{x}+3\text{y}=-2$ $-3\text{x}+3\Big(-\frac{1}2\Big)=-2$ $-3\text{x}-\frac{3}{2}=-2$ $-3\text{x}=\frac{-2}1+\frac{3}2$ $-3\text{x}=\frac{-4+3}2$ $-3\text{x}=\frac{-1}2$ $\text{x}=\frac{-1}{-6}$ $\text{x}=\frac{1}6$ Now, put the value of x and y in equation (iii), $5\text{x}-4\text{y}=3$ $5\Big(\frac{1}6\Big)-4\Big(-\frac{1}2\Big)=3$ $\frac{5}6+\frac{4}2=3$ $\frac{5+12}6=3$ $\frac{17}6=3$ $\text{LHS}\neq\text{RHS}$So, value of x and y do not satisfy the equation (iii).
So, vectors $\vec{\text{a}}-2\vec{\text{b}}+3\vec{\text{c}},\ -3\vec{\text{b}}+5\vec{\text{c}},\ -2\vec{\text{a}}+3\vec{\text{b}}-4\vec{\text{c}}$ are not coplanar.
View full question & answer→Question 605 Marks
Find the value of $\lambda$ for which the four points with position vectors
$-\hat{\text{j}}-\hat{\text{k}},4\hat{\text{i}}+5\hat{\text{j}}+\lambda\hat{\text{k}},3\hat{\text{i}}+9\hat{\text{j}}+4\hat{\text{k}}$ and $-4\hat{\text{i}}+4\hat{\text{j}}+4\hat{\text{k}}$ are co planar.
AnswerLet
position vector of $\text{A}=-\hat{\text{j}}-\hat{\text{k}}$
position vector of $\text{B}=4\hat{\text{i}}+5\hat{\text{j}}+\lambda\hat{\text{k}}$
position vector of $\text{C}=3\hat{\text{i}}+9\hat{\text{j}}+4\hat{\text{k}}$
position vector of $\text{D}=-4\hat{\text{i}}+4\hat{\text{j}}+4\hat{\text{k}}$
The four points are coplanar if the vectors $\vec{\text{AB}},\vec{\text{AC}},\vec{\text{AD}}$ are coplanar.
$\vec{\text{AB}}=4\hat{\text{i}}+6\hat{\text{j}}+(\lambda+1)\hat{\text{k}}$
$\vec{\text{AC}}=3\hat{\text{i}}+10\hat{\text{j}}+5\hat{\text{k}}$
$\vec{\text{AD}}=4\hat{\text{i}}+5\hat{\text{j}}+5\hat{\text{k}}$
$\begin{vmatrix}4&6&(\lambda+1)\\3&10&5\\-4&5&5 \end{vmatrix}=0$
$4(50-25)-6(15+20)+(\lambda+1)(15+40)=0$
$100-210+55+55\lambda=0$
$55\lambda=55$
$\lambda=1$
View full question & answer→Question 615 Marks
The two vectors $\hat{\text{j}}+\hat{\text{k}}$ and $3\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}$ represents the sides $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{AC}}$ respectively of a triangle ABC. Find the length of the median through A.
AnswerDisclaimer: The question has been solved by taking the vector $\overrightarrow{\text{AB}}$ as $\hat{\text{j}}+\hat{\text{k}}$.In $\triangle\text{ABC},\ \overrightarrow{\text{AB}}=\hat{\text{j}}+\hat{\text{k}}$ and $\overrightarrow{\text{AC}}=3\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}$
Let the position vector of A be (0, 0, 0). Then, the position vectors of B and C are (0, 1, 1) and (3, -1, 4), respectively.
Suppose D be the mid-point of the line segment joining the points B(0, 1, 1) and C(3, -1, 4).
$\therefore$ position vector of D $=\frac{\big(\hat{\text{j}}+\hat{\text{k}}\big)+\big(3\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}\big)}{2}=\frac{3\hat{\text{i}}+5\hat{\text{k}}}{2}=\frac{3}2\hat{\text{i}}+\frac{5}2\hat{\text{k}}$
Now,
Length of the median, AD =
$\Big|\overrightarrow{\text{AD}}\Big|=\Big|\Big(\frac{3}2\hat{\text{i}}+\frac{5}2\hat{\text{k}}\Big)-\big(0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}\big)\Big|$
$=\Big|\frac{3}2\hat{\text{i}}+\frac{5}2\hat{\text{k}}\Big|$
$=\sqrt{\Big(\frac{3}{2}\Big)^2+0^2+\Big(\frac{5}2\Big)^2}$
$=\sqrt{\frac{34}{4}}$
$=\sqrt{\frac{17}2}\text{units}$ View full question & answer→Question 625 Marks
If $\vec{\text{p}}$ and $\vec{\text{q}}$ are unit vectors forming an angle of 30°; find the area of the parallelogram having $\vec{\text{a}}=\vec{\text{p}}+2\vec{\text{q}}$ and $\vec{\text{b}}=2\vec{\text{p}}+\vec{\text{q}}$ as its diagonals.
AnswerGiven $\vec{\text{p}}$ and $\vec{\text{q}}$ be unit vector with angle 30° between then
$|\vec{\text{p}}|=|\vec{\text{q}}|=1$
$\vec{\text{p}}\times\vec{\text{q}}=|\vec{\text{p}}|\vec{\text{q}}|\sin30^{\circ}\hat{\text{n}}$
$=1.1\big(\frac{1}{2}\big)\hat{\text{n}}$
$|\vec{\text{p}}\times\vec{\text{q}}|=\big|\frac{\hat{\text{n}}}{2}\big|$
$\big|\vec{\text{p}}\times\vec{\text{q}}\big|=\frac{1}{2}\dots(1)$ [since, $\hat{\text{n}}$ is a unit vector]
Area of parallelogram $=\frac{1}{2}\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$=\frac{1}{2}\big|\big(\vec{\text{p}}+2\vec{\text{q}}\big)\times\big(2\vec{\text{p}}+\vec{\text{q}}\big)\big|$
$=\frac{1}{2}|\vec{\text{p}}\times2\vec{\text{p}}+\vec{\text{p}}\times\vec{\text{q}}+2\vec{\text{q}}\times2\vec{\text{p}}+2\vec{\text{q}}\times\vec{\text{q}}|$
$=\frac{1}{2}|\vec{0}+\vec{\text{p}}\times\vec{\text{q}}+2\vec{\text{q}}\times2\vec{\text{p}}+\vec{0}|$ $\big[\text{Since, }\vec{\text{p}}\times2\vec{\text{q}}=\vec{0}\text{ and }2\vec{\text{q}}\times\vec{\text{q}}=\vec{0}\big]$
$=\frac{1}{2}|\vec{\text{p}}\times\vec{\text{q}}+4\big(\vec{\text{q}}\times\vec{\text{p}}\big)|$
$=\frac{1}{2}|\big(\vec{\text{p}}\times\vec{\text{q}}\big)-4\big(\vec{\text{p}}\times\vec{\text{q}}\big)|$ $\big[\text{Since, }\vec{\text{q}}\times\vec{\text{p}}=-\vec{\text{p}}\times\vec{\text{q}}\big]$
$=\frac{1}{2}|-3\big(\vec{\text{p}}\times\vec{\text{q}}\big)|$
$=\frac{3}{2}|\vec{\text{p}}\times\vec{\text{q}}|$
$=\frac{3}{2}\times\frac{1}{2}$ [using (1)]
$=\frac{3}{4}$
Area of parallelogram $=\frac{3}{4}\text{ sq. unit}$
View full question & answer→Question 635 Marks
Show that the four points A, B, C, D with position vectors $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}},\ \vec{\text{d}}$ respectively such that $3\vec{\text{a}}-2\vec{\text{b}}+5\vec{\text{c}}-6\vec{\text{d}}=0$, are coplanar. Also, find the position vector of the point of intersection of the line segments AC and BD.
AnswerLet AC and BD intersects at a point P. We have, $3\vec{\text{a}}-2\vec{\text{b}}+5\vec{\text{c}}-6\vec{\text{d}}=0$ $\Rightarrow3\vec{\text{a}}+5\vec{\text{c}}=2\vec{\text{b}}+6\vec{\text{d}}$ Since sum of co-efficients on both sides of the above equation is 8. so we divide the equation on both sides by 8. $\Rightarrow\frac{3\vec{\text{a}}+5\vec{\text{c}}}8=\frac{2\vec{\text{b}}+6\vec{\text{d}}}8$ $\Rightarrow\frac{3\vec{\text{a}}+5\vec{\text{c}}}{3+5}=\frac{2\vec{\text{b}}+6\vec{\text{d}}}{2+6}$
Therefore, P divides AC in the ratio of 3 : 5 and P divides BD in the ratio of 2 : 6. Therefore, position vector of the point of intersection of AC and BD will be $\Rightarrow\frac{3\vec{\text{a}}+5\vec{\text{c}}}8=\frac{2\vec{\text{b}}+6\vec{\text{d}}}8$ View full question & answer→Question 645 Marks
If $\vec{\text{a}},\vec{\text{b}}$ are the position vectors of A, B respectively, find the position vector of a point C in AB produced such that AC = 3AB and that a point D in BA produced such that BD = 2BA.
Answer
Let the position vectors of C and D are $\vec{\text{c}}\text{ and }\vec{\text{d}}$ respectively. We have,
AC = 3AB
⇒ AC = 3(AC - BC)
⇒ 2AC = 3BC
$\Rightarrow\frac{\text{AC}}{\text{BC}}=\frac{3}{2}$
So, C divides AB in the ratio of 3 : 2 externally.
$\vec{\text{c}}=\frac{2\vec{\text{a}}-3\vec{\text{b}}}{2-3}=3\vec{\text{b}}-2\vec{\text{a}}$
Position vector of point C is $3\vec{\text{b}}-2\vec{\text{a}}$
Moreover,
BD = 2BA
⇒ BD = 2(BD - AD)
⇒ BD = 2 AD
$\Rightarrow\frac{\text{BD}}{\text{AD}}=\frac{2}1$
$\therefore\ \vec{\text{d}}=\frac{\vec{\text{b}}-2\vec{\text{a}}}{1-2}=2\vec{\text{a}}-\vec{\text{b}}$
Position vector of point D is $2\vec{\text{a}}-\vec{\text{b}}$ View full question & answer→Question 655 Marks
Show that four points whose position vectors are
$6\hat{\text{i}}-7\hat{\text{j}},16\hat{\text{i}}-19\hat{\text{j}}-4\hat{\text{k}},3\hat{\text{i}}-6\hat{\text{k}},2\hat{\text{i}}-5\hat{\text{j}}+10\hat{\text{k}}$ are coplanar.
AnswerDISCLAIMER: Given points are not coplaner.
Let A, B, C, D be the given points. The given points will be coplanar iff any one of the follewing triads of vectors are coplanar:
$\vec{\text{AB}},\vec{\text{AC}},\vec{\text{AD}};\vec{\text{AB}},\vec{\text{BC}},\vec{\text{CD}};\vec{\text{BC}},\vec{\text{BA}},\vec{\text{BD}}$ etc.
In order to show that $\vec{\text{AB}},\vec{\text{AC}},\vec{\text{AD}}$ are coplanar, we will have to show that their scaler triple
product i.e. $\Big[\vec{\text{AB}}\vec{\text{ AC }}\vec{\text{AD}}\Big]=0$
Using, $\vec{\text{PQ}}$ = Position vector of Q - position vector of P, we obtain
Now,
$\vec{\text{AB}}=(16\hat{\text{i}}-19\hat{\text{j}}-4\hat{\text{k}})-(6\hat{\text{i}}-7\hat{\text{j}})\\=10\hat{\text{i}}-12\hat{\text{j}}-4\hat{\text{k}}$
$\vec{\text{AC}}=(3\hat{\text{i}}-6\hat{\text{k}})-(6\hat{\text{i}}-7\hat{\text{j}})\\=-3\hat{\text{i}}+7\hat{\text{j}}-6\hat{\text{k}}$
and, $\vec{\text{AD}}=(2\hat{\text{i}}-5\hat{\text{j}}+10\vec{\text{k}})-(6\hat{\text{i}}-7\hat{\text{j}})\\=-4\hat{\text{i}}+2\hat{\text{j}}+10\hat{\text{k}}$
$\therefore\Big[\vec{\text{AC}}\vec{\text{ AC }}\vec{\text{AD}}\Big]=\begin{vmatrix}10&-12&-4\\-3&7&-6\\-4&2&10 \end{vmatrix}$
$=10(70+12)+12(-30-24)-4(-6+28)=84$
Thus, the given points are not coplanar.
View full question & answer→Question 665 Marks
Prove that the points having position vectors $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\ 3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}$ and $-3\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}$ are collinear.
AnswerLet A, B, C be the points with position vectors $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\ 3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}$ and $-3\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}$
Then,
$\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A
$=3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}-\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
$=2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$
$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=-3\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}-3\hat{\text{i}}-4\hat{\text{j}}-7\hat{\text{k}}$
$=-6\hat{\text{i}}-6\hat{\text{j}}-12\hat{\text{k}}$
$=-3\big(2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}\big)$
$\therefore\ \overrightarrow{\text{BC}}=-3\overrightarrow{\text{AB}}$
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are parallel vectors.
But B is a point common to them.
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are collinear.
Hence, A, B, C are collinear.
View full question & answer→Question 675 Marks
$\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are the position vectors of points A, B and C respectively, prove that:
$\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}$ is a vector perpendicular to the plane of triangle ABC.
AnswerWe know that if any vector is perpendicular to all three sides of $\triangle\text{ABC},$ it must be perpendicular to the plane of $\triangle \text{ABC}.$
Now,
$\vec{\text{AB}}=\vec{\text{b}}-\vec{\text{a}},\vec{\text{BC}}=\vec{\text{c}}-\vec{\text{b}},\vec{\text{CA}}=\vec{\text{a}}-\vec{\text{c}}$
$\big($position vectors of A, B and C are $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}\big)$
We have
$\vec{\text{AB}}.\big(\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\big(\vec{\text{b}}-\vec{\text{a}}\big).\big(\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\vec{\text{b}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)+\vec{\text{b}}\big(\vec{\text{b}}\times\vec{\text{c}}\big)+\vec{\text{b}}\big(\vec{\text{c}}\times\vec{\text{a}}\big)\\-\vec{\text{a}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)-\vec{\text{a}}\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\vec{\text{a}}.\big(\vec{\text{c}}\times\vec{\text{a}}\big)$ (By distributive law)
$=\big[\vec{\text{b}}\ \vec{\text{a}}\ \vec{\text{b}}\big]+\big[\vec{\text{b}}\ \vec{\text{b}}\ \vec{\text{c}}\big]+\big[\vec{\text{b}}\ \vec{\text{c}}\ \vec{\text{a}}\big]-\big[\vec{\text{a}}\ \vec{\text{a}}\ \vec{\text{b}}\big]-\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]-\big[\vec{\text{a}}\ \vec{\text{c}}\ \vec{\text{a}}\big]$
$=0+0+\big[\vec{\text{b}}\ \vec{\text{c}}\ \vec{\text{a}}\big]-0-\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]-0$
$=0\ \big(\therefore\big[\vec{\text{b}}\ \vec{\text{c}}\ \vec{\text{a}}\big]=\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]\big)$
$\vec{\text{BC}}\big(\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\big(\vec{\text{c}}-\vec{\text{b}}\big).\big(\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\vec{\text{c}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)+\vec{\text{c}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)+\vec{\text{c}}.\big(\vec{\text{c}}\times\vec{\text{a}}\big)\\-\vec{\text{b}}\big(\vec{\text{a}}\times\vec{\text{b}}\big)-\vec{\text{b}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\vec{\text{b}}.\big(\vec{\text{c}}\times\vec{\text{a}}\big)$ (By distributive law)
$=\big[\vec{\text{c}}\ \vec{\text{a}}\ \vec{\text{b}}\big]+\big[\vec{\text{c}}\ \vec{\text{b}}\ \vec{\text{c}}\big]+\big[\vec{\text{c}}\ \vec{\text{c}}\ \vec{\text{a}}\big]-\big[\vec{\text{b}}\ \vec{\text{a}}\ \vec{\text{b}}\big]-\big[\vec{\text{b}}\ \vec{\text{b}}\ \vec{\text{c}}\big]-\big[\vec{\text{b}}\ \vec{\text{c}}\ \vec{\text{a}}\big]$
$=\big[\vec{\text{c}}\ \vec{\text{a}}\ \vec{\text{b}}\big]+0+0-0-0-\big[\vec{\text{b}}\ \vec{\text{c}}\ \vec{\text{a}}\big]$
$=0$ $=0 \ \big(\therefore\big[\vec{\text{c}}\ \vec{\text{a}}\ \vec{\text{b}}=\big[\vec{\text{b}}\ \vec{\text{c}}\ \vec{\text{a}}\big]\big)$
Similarly,
$\vec{\text{CA}}.\big(\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\big(\vec{\text{a}}-\vec{\text{c}}\big).\big(\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\vec{\text{a}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)+\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)+\vec{\text{a}}\big(\vec{\text{c}}\times\vec{\text{a}}\big)\\-\vec{\text{c}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)-\vec{\text{c}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\vec{\text{c}}.\big(\vec{\text{c}}\times{\vec{\text{a}}}\big)$ (By distributive law)
$=\big[\vec{\text{a}}\vec{\text{a}}\vec{\text{b}}\big]+\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]+\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{a}}\big]-\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]-\big[\vec{\text{c}}\vec{\text{b}}\vec{\text{c}}\big]-\big[\vec{\text{c}}\vec{\text{c}}\vec{\text{a}}\big]$
$=0\ \big(\therefore\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]=\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]\big)$
Hence, vector $\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}$ is
perpendicular to all sides of $\triangle\text{ABC}$ and also perpendicular to the plane of $\triangle\text{ABC}.$
View full question & answer→Question 685 Marks
Prove that a necessary and sufficient condition for three vectors $\vec{\text{a}},\ \vec{\text{b}}$ and $\vec{\text{c}}$ to be coplanar is that there exist scalars l, m, n not all zero simultaneously such that $\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=\vec0$.
AnswerNecessary Condition: Let $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are three coplanar vectors. Then one of them can be expressed as the linear combination of other two vectors. Let, $\vec{\text{c}}=\text{x}\vec{\text{a}}+\text{y}\vec{\text{b}}$ $\text{x}\vec{\text{a}}+\text{y}\vec{\text{b}}-\vec{\text{c}}=0$ Put $\text{x}=1,\ \text{y}=\text{m},\ (-1)=\text{n}$$\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=0$
Thus, if $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are coplanar vectors, then there exist scalars l, m, n $\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=0$ Such that l, m, n are not all zero simultaneously. Sufficient Condition: Let $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ be three vectors such that there exist scalars l, m, n not all zero simultaneously satisfying $\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=0$ $\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=0$ $\text{n}\vec{\text{c}}=-\text{l}\vec{\text{a}}-\text{m}\vec{\text{b}}$ Dividing by n, both the sides $\frac{\text{n}\vec{\text{c}}}{\text{n}}=\frac{-\text{l}\vec{\text{a}}}{\text{n}}-\frac{\text{m}\vec{\text{b}}}{\text{n}}$ $\vec{\text{c}}=\Big(-\frac{\text{l}}{\text{n}}\Big)\vec{\text{a}}+\Big(-\frac{\text{m}}{\text{n}}\Big)\vec{\text{b}}$ $\vec{\text{c}}$ is a linear combination of $\vec{\text{a}} \text{ and }\vec{\text{b}}$ Hence, $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are coplanar vectors.
View full question & answer→Question 695 Marks
A unit vector $\vec{\text{a}}$ makes angles $\frac{\pi}{4}$ and $\frac{\pi}{3}$ with $\hat{\text{i}}$ and $\hat{\text{j}}$ respectively and an acute angle $\theta$ with $\hat{\text{k}}$. find the angle $\theta$ and components of $\vec{\text{a}}$ .
AnswerLet $\vec{\text{a}}=\text{a}_{1}\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}},$ where $\text{a}_1,\text{a}_2$ and $\text{a}_3$ are components of $\vec{\text{a}.}$
$\Rightarrow{\text{a}_1}^2+{\text{a}_2}^2+{\text{a}_3}^2=1$ (Because $\vec{\text{a}}$ is a unit vector) ...(1)
Now,
$\vec{\text{a}}.\hat{\text{i}}=\text{a}_1$
$\Rightarrow|\vec{\text{a}}||\hat{\text{i}}|\cos\frac{\pi}{4}=\text{a}_1$ (Because the angle between $\vec{\text{a}}$ and $\hat{\text{i}}$ is $\frac{\pi}{4}$)
$\Rightarrow(1)(1)\frac{1}{\sqrt{2}}=\text{a}_1$ (Because $\vec{\text{a}}$ and $\hat{\text{i}}$ are unit vectors)
$\Rightarrow \text{a}_1=\frac{1}{\sqrt{2}}$
Again,
$\vec{\text{a}}.\hat{\text{j}}=\text{a}_2$
$\Rightarrow|\vec{\text{a}}||\hat{\text{i}}|\cos\frac{\pi}{3}=\text{a}_2$ (Becasue the angle between $\vec{\text{a}}$ and $\hat{\text{i}}$ is $\frac{\pi}{3}$)
$\Rightarrow(1)(1)\frac{1}{2}=\text{a}_2$ (Because $\vec{\text{a}}$ and $\hat{\text{i}}$ are unite vectors)
$\Rightarrow{\text{a}}_2=\frac{1}{2}$
Now from (1),
$\Big(\frac{1}{\sqrt{2}}\Big)^2+\big(\frac{1}{2}\big)^2+{\text{a}_3}^2=1$
View full question & answer→Question 705 Marks
If $\vec{\alpha}=3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}$ and $\vec{\beta}=2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}},$ then express $\vec{\beta}$ in the form of $\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2,$ where $\vec{\beta}_1$ is parallel to $\vec{\alpha}$ and $\vec{\beta}_2$ is perpendicular to $\vec{\alpha}$.
AnswerGiven that $\vec{\alpha}=3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}$ and $\vec{\beta}=2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}$
Also,
$\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2,$
$\Rightarrow\vec{\beta}_2=\vec{\beta}+\vec{\beta}_1\dots(1)$
Since $\vec{\beta}_1$ is parallel to $\vec{\alpha},$
$\vec{\beta}_1=\text{t}\vec{\alpha}$
$\Rightarrow\vec{\beta}_1=\text{t}\big(3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}\big)=3\text{t}\hat{\text{i}}+4\text{t}\hat{\text{j}}+5\text{t}\hat{\text{k}}\dots(2)$
Substituting the values of $\vec{\beta}_1$ and $\vec{\alpha}$ in (1),we get
$\vec{\beta}_2=2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}-\big(3\text{t}\hat{\text{i}}+4\text{t}\hat{\text{j}}+5\text{t}\hat{\text{k}}\big)\\=(2-3\text{t})\hat{\text{i}}+(1-4\text{t})\hat{\text{j}}+(-4-5\text{t})\hat{\text{k}}\dots(3)$Since $\vec{\beta}_2$ is perpendicular to $\vec{\alpha},$
$\vec{\beta}_2.\vec{\alpha}=0$
$\Rightarrow\Big[(2-3\text{t})\hat{\text{i}}+(1-4\text{t})\hat{\text{j}}+(-4-5\text{t})\hat{\text{k}}\Big].\big(3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}\big)=0$
$\Rightarrow3(2-3\text{t})+4(1-4\text{t})+5(-4-5\text{t})=0$
$\Rightarrow6-9\text{t}+4-16\text{t}-20-25\text{t}=0$
$\Rightarrow-50\text{t}=10$
$\Rightarrow\text{t}=\frac{-1}{5}$
From (2) and (3), we get
$\vec{\beta}_1=\frac{-1}{5}\big(3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}\big)$
$\vec{\beta}_2=\frac{13}{5}\hat{\text{i}}+\frac{9}{5}\hat{\text{j}}-3\hat{\text{k}}=\frac{1}{5}\big(13\hat{\text{i}}+9\hat{\text{j}}-15\hat{\text{k}}\big)$
View full question & answer→Question 715 Marks
ABCD are four points in a plane and Q is the point of intersection of the lines joining the mid-points of AB and CD, BC and AD. Show that $\overrightarrow{\text{PA}}+\overrightarrow{\text{PB}}+\overrightarrow{\text{PC}}+\overrightarrow{\text{PD}}=4\ \overrightarrow{\text{PQ}}$, where P ia any point.
Answer
Let $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}},\ \vec{\text{d}}$ be the position vectors of the points A, B, C and D respectively.
Then, position vector of
mid-point of $\text{AB}=\frac{\vec{\text{a}}+\vec{\text{b}}}2$
mid-point of $\text{BC}=\frac{\vec{\text{b}}+\vec{\text{c}}}2$
mid-point of $\text{CD}=\frac{\vec{\text{c}}+\vec{\text{d}}}2$
mid-point of $\text{DA}=\frac{\vec{\text{a}}+\vec{\text{d}}}2$
Q is the mid-point of the line joining the mid-points of AB and CD
$\therefore\ \text{Q}=\frac{\frac{\vec{\text{a}}+\vec{\text{a}}}2+\frac{\vec{\text{c}}+\vec{\text{d}}}2}{2}$
$=\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}+\vec{\text{d}}}4$
Let $\vec{\text{p}}$ be the position vector of P.
Then,
$\overrightarrow{\text{PA}}+\overrightarrow{\text{PB}}+\overrightarrow{\text{PC}}+\overrightarrow{\text{PD}}$
$=\vec{\text{a}}-\vec{\text{p}}+\vec{\text{b}}-\vec{\text{p}}+\vec{\text{c}}-\vec{\text{p}}+\vec{\text{d}}-\vec{\text{p}}$
$=\big(\vec{\text{a}}+\vec{\text{p}}+\vec{\text{c}}+\vec{\text{d}}\big)-4\vec{\text{p}}$
$=4\bigg(\frac{\vec{\text{a}}+\vec{\text{p}}+\vec{\text{c}}+\vec{\text{d}}}{4}-\vec{\text{p}}\bigg)$
$=4\ \overrightarrow{\text{PQ}}$ View full question & answer→Question 725 Marks
Show that the vector $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ is equally inclined to the coordinate axes.
AnswerLet $\theta_1$ be the angle between $\vec{\text{a}}$ and x-axis.
$|\vec{\text{a}}|=\sqrt{(1)^2+(1)^2+(1)^2}=\sqrt{3}$
$\vec{\text{b}}=\hat{\text{i}}$ (Because $\hat{\text{i}}$ is the unit vector along x-axis)
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2}=\sqrt{1}=1$
$\vec{\text{a}}.\vec{\text{b}}=1+0+0=1$
$\cos\theta_{1}=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{a}|\big|\vec{\text{b}}\big|}=\frac{1}{(\sqrt{3})(1)}=\frac{1}{\sqrt{3}}$
$\Rightarrow\theta_{1}=\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)\dots(1)$
Let $\theta_{2}$ be the angle between $\vec{\text{a}}$ and y-axis.
$|\vec{\text{a}}|=\sqrt{(1)^2+(1)^2+(1)^2}=\sqrt{3}$
$\vec{\text{b}}=\hat{\text{j}}$ (Because $\hat{\text{j}}$ is the unit vector along y-axis)
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2}=\sqrt{1}=1$
$\vec{\text{a}}.\vec{\text{b}}=0+1+0=1$
$\cos\theta_{2}=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{a}|\big|\vec{\text{b}}\big|}=\frac{1}{(\sqrt{3})(1)}=\frac{1}{\sqrt{3}}$
$\Rightarrow\theta_{2}=\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)\dots(2)$
Let $\theta_{3}$ be the angle between $\vec{\text{a}}$ and z-axis.
$|\vec{\text{a}}|=\sqrt{(1)^2+(1)^2+(1)^2}=\sqrt{3}$
$\vec{\text{b}}=\hat{\text{k}}$ (Because $\hat{\text{k}}$ is the unit vector along z-axis)
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2}=\sqrt{1}=1$
$\vec{\text{a}}.\vec{\text{b}}=0+0+1=1$
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{a}|\big|\vec{\text{b}}\big|}=\frac{1}{(\sqrt{3})(1)}=\frac{1}{\sqrt{3}}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)\dots(3)$
From (1), (2) and (3), the given vector is equally inclined to the coordinate axes.
View full question & answer→Question 735 Marks
In a triangle OAB, $\angle\text{AOB}=90^\circ.$ If P and Q are points of trisection of AB, prove that $\text{OP}^2+\text{OQ}^2=\frac{5}{9}\text{AB}^2.$
AnswerLet $\vec{\text{o}},\vec{\text{a}}$ and $\vec{\text{b}}$ be the position vector of the O,A and B.
P and Q are points of trisection of AB.
Position vector of point $\text{P}= \frac{2\vec{\text{a}}+\vec{\text{b}}}{3}$
Position vector of point $\text{Q}= \frac{\vec{\text{a}}+2\vec{\text{b}}}{3}$
$\text{OP}=\frac{2\vec{\text{a}}+\vec{\text{b}}}{3}-\vec{\text{O}}=\frac{2\vec{\text{a}}+\vec{\text{b}}-3\vec{\text{o}}}{3}=\frac{20\text{A+OB}}{3}$
$\text{OQ}=\frac{\vec{\text{a}}+2\vec{\text{b}}}{3}-\vec{\text{O}}=\frac{\vec{\text{a}}+2\vec{\text{b}}-3\vec{\text{o}}}{3}=\frac{\text{OA+2OB}}{3}$
$\text{OP}^2+\text{OQ}^2=\Big(\frac{2\text{OA}+\text{OB}}3\Big)^{2}+\Big(\frac{\text{OA}+2\text{OB}}{3}\Big)^{2}$
$\frac{5(\text{OA}^{2}+\text{OB}^{2})+8(\text{OA})(\text{OB})\cos90^{\circ}}{9}$
$=\frac{5}{9}\text{AB}^{2}\dots \big[\because \text{OA}^{2}+\text{OB}^{2}=\text{AB}^{2} \text{and}\cos90^{\circ}=0\big]$
View full question & answer→Question 745 Marks
If $|\vec{\text{a}}|=13,\big|\vec{\text{b}}\big|=5$ and $\vec{\text{a}}.\vec{\text{b}}=60,$ then find $\big|\vec{\text{a}}\times\vec{\text{b}}\big|.$
AnswerWe know that, if $\theta$ is angle between $\vec{\text{a}}$ and $\vec{\text{b}},$
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}||\vec{\text{b}}|\cos\theta$
$60=13.5.\cos\theta$
$\cos\theta=\frac{60}{65}$
$\cos\theta=\frac{12}{13}$
$\sin^2\theta=1-\cos^2\theta$
$=1-\Big(\frac{12}{13}\Big)^2$
$=1-\frac{144}{169}$
$=\frac{169-144}{169}$
$=\frac{25}{169}$
$\sin\theta=\pm\sqrt{\frac{25}{169}}$
$=\pm\frac{5}{13}$
$|\sin\theta|=\frac{5}{13}$
We know that,
$\vec{\text{a}}\times\vec{\text{b}}=|\vec{\text{a}}|.\big|\vec{\text{b}}\big|.\sin\theta.\hat{\text{n}}$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|.\big|\vec{\text{b}}\big|.|\sin\theta|.|\hat{\text{n}}|$
$=13.5.\frac{5}{13}.1$ [Since, $\hat{\text{n}}$ is aunit vector]
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=25$
View full question & answer→Question 755 Marks
Show that the points A(1, -2, -8), B(5, 0, -2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.
AnswerThe given points are A(1, -2, -8), B(5, 0, -2), and C(11, 3, 7).
$\therefore\overrightarrow{\text{AB}}=(5-1)\hat{\text{i}}+(0+2)\hat{\text{j}}+(-2+8)\hat{\text{k}}$ $=4\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$
$\overrightarrow{\text{BC}}=(11-5)\hat{\text{i}}+(3-0)\hat{\text{j}}+(7+2)\hat{\text{k}}$ $=6\hat{\text{i}}+3\hat{\text{j}}+9\hat{\text{k}}$
$\overrightarrow{\text{AC}}=(11-1)\hat{\text{i}}+(3+2)\hat{\text{j}}+(7+8)\hat{\text{k}}$ $=10\hat{\text{i}}+5\hat{\text{j}}+15\hat{\text{k}}$
$\Big|\overrightarrow{\text{AB}}\Big|=\sqrt{4^2+2^2+6^2}=\sqrt{16+4+36}$ $=\sqrt{56}=2\sqrt{14}$
$\Big|\overrightarrow{\text{BC}}\Big|=\sqrt{6^2+3^2+9^2}=\sqrt{36+9+81}$ $=\sqrt{126}=3\sqrt{14}$
$\Big|\overrightarrow{\text{AC}}\Big|=\sqrt{10^2+5^2+15^2}=\sqrt{100+25+225}$ $=\sqrt{350}=5\sqrt{14}$
$\therefore\Big|\overrightarrow{\text{AC}}\Big|=\Big|\overrightarrow{\text{AB}}\Big|+\Big|\overrightarrow{\text{BC}}\Big|$
Thus, the given points A, B, and C are collinear.
Now, let point B divide AC in the ratio $\lambda:1.$ Then we have:
$\overrightarrow{\text{OB}}=\frac{\lambda\overrightarrow{\text{OC}}+\overrightarrow{\text{OA}}}{(\lambda+1)}$
$\Rightarrow5\hat{\text{i}}-2\hat{\text{k}}=\frac{\lambda\big(11\hat{\text{i}}+3\hat{\text{j}}+7\hat{\text{k}}\big)+\big(\hat{\text{i}-2\hat{\text{j}}-8\hat{\text{k}}}\big)}{\lambda+1}$
$\Rightarrow{(\lambda+1)}\big(5\hat{\text{i}}-2\hat{\text{k}}\big)$ $=11\lambda\hat{\text{i}}+3\lambda\hat{\text{j}}+\hat{\text{i}}-2\hat{\text{j}}-8\hat{\text{k}}$
$\Rightarrow5{(\lambda+1)}\hat{\text{i}}-2(\lambda+1)\hat{\text{k}}$ $=(11\lambda+1)\hat{\text{i}}+(3\lambda-2)\hat{\text{j}}+(7\lambda-8)\hat{\text{k}}$
On equating the corresponding components, we get:
$5(\lambda+1)=11\lambda+1$
$\Rightarrow5{\lambda+5}=11\lambda+1$
$\Rightarrow6{\lambda}=4$
$\Rightarrow{\lambda}=\frac{4}{6}=\frac{2}{3}$
Hence, point B divides AC in the ratio 2 : 3
View full question & answer→Question 765 Marks
Prove that in any triangle ABC, $\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}},$ where a, b, c are the magnitudes of the sides opposite to the vertices A, B, C, respectively.
AnswerLet, $\overrightarrow{\text{AB}}=\text{c},\overrightarrow{\text{BC}}=\text{a}$ and $\overrightarrow{\text{AC}}=\text{b}$
Hare, components of c are ccos A and csin A is drawn.
Since, $\overrightarrow{\text{CD}}=\text{b}-\text{c}\cos\text{A}$
In $\triangle\text{BDC},$
$\text{a}^2=(\text{b}-\text{c}\cos\text{A})^2+(\text{c}\sin\text{A})^2$
$\Rightarrow\text{a}^2=\text{b}^2+\text{c}^2\cos^2\text{A}-2\text{bc}\cos\text{A}+\text{c}^2\sin^2\text{A}$
$\Rightarrow2\text{b}\text{c}\cos\text{A}=\text{b}^2-\text{a}^2+\text{c}^2(\cos^2\text{A}+\sin^2\text{A})$
$\therefore\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}$ View full question & answer→Question 775 Marks
Let $\vec{\text{u}},\vec{\text{v}}$ and $\vec{\text{w}}$ be vectors such $\vec{\text{u}}+\vec{\text{v}}+\vec{\text{w}}=\vec{0}.$ If $|\vec{\text{u}}|=3,|\vec{\text{v}}|=4$ and $|\vec{\text{w}}|=5,$ then find $\vec{\text{u}}.\vec{\text{v}}+\vec{\text{v}}.\vec{\text{w}}+\vec{\text{w}}.\vec{\text{u}}.$
AnswerHere, $\vec{\text{u}}+\vec{\text{v}}+\vec{\text{w}}=0$
Squaring both the sides,
$\big(\vec{\text{u}}+\vec{\text{v}}+\vec{\text{w}}\big)^2=(0)^2$
$|\vec{\text{u}}|^2+|\vec{\text{v}}|^2+|\vec{\text{w}}|^2+2\vec{\text{u}}\vec{\text{v}}+2\vec{\text{v}}\vec{\text{w}}+2\vec{\text{w}}\vec{\text{u}}=0$
$(3)^2+(4)^2+(5)^2+2\big(\vec{\text{u}}.\vec{\text{v}}+\vec{\text{v}}.\vec{\text{w}}+\vec{\text{w}}.\vec{\text{u}}\big)=0$
$9+16+25+2\big(\vec{\text{u}}\vec{\text{v}}+\vec{\text{v}}\vec{\text{w}}+\vec{\text{w}}\vec{\text{u}}\big)=0$
$2\big(\vec{\text{u}}\vec{\text{v}}+\vec{\text{v}}\vec{\text{w}}+\vec{\text{w}}\vec{\text{u}}\big)=-50$
$\vec{\text{u}}\vec{\text{v}}+\vec{\text{v}}\vec{\text{w}}+\vec{\text{w}}\vec{\text{u}}=\frac{-50}{2}$
$\vec{\text{u}}\vec{\text{v}}+\vec{\text{v}}\vec{\text{w}}+\vec{\text{w}}\vec{\text{u}}=-25$
View full question & answer→Question 785 Marks
Using vectors, find the area of the triangle ABC with vertices A(1, 2, 3), B(2, –1, 4) and C(4, 5, –1).
AnswerHere, $\overrightarrow{\text{AB}}=(2-1)\hat{\text{i}}+(-1-2)\hat{\text{j}}+(4-3)\hat{\text{k}}$
$=\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}$
and $\overrightarrow{\text{AC}}=(4-1)\hat{\text{i}}+(5-2)\hat{\text{j}}+(-1-3)\hat{\text{k}}$
$=3\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}}$
$\therefore\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}} \\1&-3&1\\3&3&-4 \end{vmatrix}$
$=\hat{\text{i}}(12-3)-\hat{\text{j}}(-4-3)+\hat{\text{k}}(3+9)$
$=9\hat{\text{i}}+7\hat{\text{j}}+12\hat{\text{k}}$
and $|\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}|=\sqrt{9^2+7^2+12^2}$
$=\sqrt{81+49+144}$
$=\sqrt{274}$
$\therefore\text{Area of }\triangle\text{ABC}=\frac{1}{2}|\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}|$
$=\frac{1}{2}\sqrt{274}\text{ sq units}$ View full question & answer→Question 795 Marks
Show that the line segments joining the mid-points of opposite sides of a quadrilateral bisects each other.
AnswerLet ABCD is the quadrilateral and P, Q, R, S are mid points of the sides AB, BC, CD, DA respectively.
Join DB to form triangle ABD. $\frac{\text{AS}}{\text{SD}}=\frac{\text{AP}}{\text{PB}}$ $\Rightarrow\text{SP}||\text{DB}\text{ and }\text{SP}=\frac{1}2\text{DB}$ In triangle BCD $\frac{\text{CR}}{\text{RD}}=\frac{\text{CQ}}{\text{QB}}$ $\Rightarrow\text{RQ}||\text{DB}\text{ and }\text{RQ}=\frac{1}2\text{DB}$ In quadrilateral PQRS, SP = RQ and SP || RQ $\therefore$ PQRS is a parallelogram. Diagonals of a parallelogram bisect each other. $\therefore$ PR and QS bisect each other. View full question & answer→Question 805 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two non-collinear unit vectors such that $\big|\vec{\text{a}}+\vec{\text{b}}\big|=\sqrt{3},$ find $\big(2\vec{\text{a}}-5\vec{\text{b}}\big).\big(3\vec{\text{a}}+\vec{\text{b}}\big).$
AnswerGiven
$\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors
then, $|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=1$
$\big|\vec{\text{a}}+\vec{\text{b}}\big|=\sqrt{3}$
Squaring both the sides,
$\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=(\sqrt{3})^2$
$|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}=3$
$1+1+2\vec{\text{a}}.\vec{\text{b}}=3$
$2+2\vec{\text{a}}.\vec{\text{b}}=3$
$2\vec{\text{a}}.\vec{\text{b}}=3-2$
$2\vec{\text{a}}.\vec{\text{b}}=1$
$2\vec{\text{a}}.\vec{\text{b}}=\frac{1}{2}$
View full question & answer→Question 815 Marks
Decompose the vector $6\hat{\text{i}}-3\hat{\text{j}}-6\hat{\text{k}}$ into vectors which are parallal and perpendicular to the vector $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}.$
AnswerLet $\vec{\text{a}}=6\hat{\text{i}}-3\hat{\text{j}}-6\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
and $\vec{\text{x}}$ and $\vec{\text{y}}$ be such that
$\vec{\text{a}}=\vec{\text{x}}+\vec{\text{y}}$
$\Rightarrow\vec{\text{y}}=\vec{\text{a}}-\vec{\text{x}}\dots(1)$
Since $\vec{\text{x}}$ is parallel to $\vec{\text{b}},$
$\vec{\text{x}}=\text{t}\vec{\text{b}}$
$\Rightarrow\vec{\text{x}}=\text{t}\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=\text{t}\hat{\text{i}}+\text{t}\hat{\text{j}}+\text{t}\hat{\text{k}}\dots(2)$
Substituting the values of $\vec{\text{x}}$ and $\vec{\text{a}}$ in (1), we get
$\vec{\text{y}}=6\hat{\text{i}}-3\hat{\text{j}}-6\hat{\text{k}}-\big(\text{t}\hat{\text{i}}+\text{t}\hat{\text{j}}+\text{t}\hat{\text{k}}\big)=(6-\text{t})\hat{\text{i}}+(-3-\text{t})\hat{\text{j}}+(-6-\text{t})\hat{\text{k}}\dots(3)$
Since $\vec{\text{y}}$ is perpendicular to $\vec{\text{b}},$
$\vec{\text{y}}.\vec{\text{b}}=0$
$\Rightarrow\big[(6-\text{t})\hat{\text{i}}+(-3-\text{t})\hat{\text{j}}+(-6-\text{t})\hat{\text{k}}\big].\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=0$
$\Rightarrow1(6-\text{t})+1(-3-\text{t})+1(-6-\text{t})=0$
$\Rightarrow-3-3\text{t}=0$
$\Rightarrow\text{t}=-1$
From (2) and (3), we get
$\vec{\text{x}}=-\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{y}}=7\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}$
So,
$\vec{\text{a}}=\vec{\text{x}}+\vec{\text{y}}=\big(-\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)+\big(7\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}\big)$
View full question & answer→Question 825 Marks
If $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}},\vec{\text{b}}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ and $\vec{\text{c}}=\text{c}_1\hat{\text{i}}+\text{c}_2\hat{\text{j}}+\text{c}_3\hat{\text{k}},$ then verify that $\vec{\text{a}}\times\big(\vec{\text{b}}+\vec{\text{c}}\big)=\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{a}}\times\vec{\text{c}}.$
AnswerWe have,
$\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$\vec{\text{b}}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$
$\vec{\text{c}}=\text{c}_1\hat{\text{i}}+\text{c}_2\hat{\text{j}}+\text{c}_3\hat{\text{k}},$
$\big(\vec{\text{b}}+\vec{\text{c}}\big)=(\text{b}_1+\text{c}_1)\hat{\text{i}}+(\text{b}_2+\text{c}_2)\hat{\text{j}}+(\text{b}_3+\text{c}_3)\hat{\text{k}}$
Now, $\vec{\text{a}}\times\big(\vec{\text{b}}+\vec{\text{c}}\big)=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1+\text{c}_{1}&\text{b}_2+\text{c}_2&\text{b}_3+\text{c}_3 \end{vmatrix}$
$=\hat{\text{i}}\big[\text{a}_2(\text{b}_3+\text{c}_3)-\text{a}_3(\text{b}_2+\text{c}_2)\big]-\hat{\text{j}}\big[\text{a}_1(\text{b}_3+\text{c}_3)-\text{a}_3(\text{b}_1+\text{c}_1)\big]\\+\hat{\text{k}\big[\text{a}}_1(\text{b}_2+\text{c}_2)-\text{a}_2(\text{b}_1+\text{c}_1)\big]$
$=\hat{\text{i}}\big[\text{a}_2\text{b}_3+\text{a}_2\text{c}_3-\text{a}_3\text{b}_2-\text{a}_3\text{c}_2\big]+\hat{\text{j}}\big[-\text{a}_1\text{b}_3-\text{a}_1\text{c}_3+\text{a}_3\text{b}_1+\text{a}_3\text{c}_1\big]\\+\hat{\text{k}}\big[\text{a}_1\text{b}_2+\text{a}_1\text{c}_2-\text{a}_2\text{b}_1-\text{a}_2\text{c}_1\big]\dots(1)$
$$$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3 \end{vmatrix}$
$=\hat{\text{i}}\big[\text{a}_2\text{b}_3-\text{a}_3\text{b}_2\big]+\hat{\text{j}}\big[\text{b}_1\text{a}_3-\text{a}_1\text{b}_3\big]+\hat{\text{k}}\big[\text{a}_1\text{b}_2-\text{a}_2\text{b}_1\big]\ \dots(2)$
$\vec{\text{a}}\times\vec{\text{c}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{a}_2&\text{a}_3\\\text{c}_1&\text{c}_2&\text{c}_3 \end{vmatrix}$
$=\hat{\text{i}}\big[\text{a}_2\text{c}_3-\text{a}_3\text{c}_2\big]+\hat{\text{j}}\big[\text{a}_3\text{c}_1-\text{a}_1\text{c}_3\big]+\hat{\text{k}}\big[\text{a}_1\text{c}_2-\text{a}_3\text{c}_1\big]\ \dots(3)$
View full question & answer→Question 835 Marks
If $\big|\vec{\text{a}}+\vec{\text{b}}\big|=60,\big|\vec{\text{a}}-\vec{\text{b}}\big|=40$ and $\big|\vec{\text{b}}\big|=46,$ find $|\vec{\text{a}}|$
AnswerHere, $\big|\vec{\text{a}}+\vec{\text{b}}\big|=60$
Squaring both the sides,
$\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=(60)^2$
$(\vec{\text{a}}+\vec{\text{b}})=(60)^2$
$(\vec{\text{a}})^2+(\vec{\text{b}})^2+2\vec{\text{a}}\vec{\text{b}}=3600$
$|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}\vec{\text{b}}=3600\dots(1)$
Now, $\big|\vec{\text{a}}-\vec{\text{b}}\big|=40$
Squaring both the sides,
$\big|\vec{\text{a}}-\vec{\text{b}}\big|^2=(40)^2$
$|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}\vec{\text{b}}=1600\dots(2)$
Adding (1) and (2),
$2|\vec{\text{a}}|^2+2\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}\vec{\text{b}}-2\vec{\text{a}}\vec{\text{b}}=3600-1600$
$2|\vec{\text{a}}|^2+2(46)^2=5200$
$2|\vec{\text{a}}|^2=5200-4232$
$2|\vec{\text{a}}|^2=968$
$|\vec{\text{a}}|^2=\frac{968}{2}$
$|\vec{\text{a}}|^2=484$
$|\vec{\text{a}}|=\sqrt{484}$
$|\vec{\text{a}}|=22$
View full question & answer→Question 845 Marks
If $\vec{\text{a}}$ be the position vector whose tip is (5, -3), find the coordinates of a point B such that $\overrightarrow{\text{AB}}=\vec{\text{a}}$, the coordinates of A being (4, -1).
AnswerLet O be the origin and let P(5, -3) be the tip of the position vector $\vec{\text{a}}$. Then,
$\vec{\text{a}}=\overrightarrow{\text{OP}}=5\hat{\text{i}}-3\hat{\text{j}}$. Let the coordinate of B be (x, y) and A has coordinates (4, -1).
Therefore,
$\overrightarrow{\text{AB}}$ = Position vector of B - Position vector of A
$=\big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}\big)-\big(4\hat{\text{i}}-\hat{\text{j}}\big)$
$=(\text{x}-4)\hat{\text{i}}+(\text{y}+1)\hat{\text{j}}$
Now,
$\overrightarrow{\text{AB}}=\vec{\text{a}}$
$\Rightarrow(\text{x}-4)\hat{\text{i}}+(\text{y}+1)\hat{\text{j}}=5\hat{\text{i}}-3\hat{\text{j}}$
$\Rightarrow\text{x}-4=5\text{ and }\text{y}+1=-3$
$\Rightarrow\text{x}=9 \text{ and }\text{y}=-4$
Hence, the coordinates of B are (9, -4).
View full question & answer→Question 855 Marks
The position vectors of points A, B and C are $\lambda\hat{\text{i}}+3\hat{\text{j}},12\hat{\text{i}}+\mu\hat{\text{j}}\text{ and }11\hat{\text{i}}-3\hat{\text{j}}$ respectively. If C divides the line segment joining A and B in the ratio 3:1, find the value of $\lambda\text{ and }\mu$
AnswerThe position vectors of points A, B and C are $\lambda\hat{\text{i}}+3\hat{\text{j}},12\hat{\text{i}}+\mu\hat{\text{j}}\text{ and }11\hat{\text{i}}-3\hat{\text{j}}$, respectively.
It is given that, C divides the line segment joining A and B in the ratio 3 : 1.
$11\hat{\text{i}}-3\hat{\text{j}}=\frac{3\times\big(12\hat{\text{i}}+\mu\hat{\text{j}}\big)+1\times\big(\lambda\hat{\text{i}}+3\hat{\text{j}}\big)}{3+1}$
$\Rightarrow11\hat{\text{i}}-3\hat{\text{j}}=\frac{(36+\lambda)\hat{\text{i}}+(3\mu+3)\hat{\text{j}}}{4}$
$\Rightarrow44\hat{\text{i}}-12\hat{\text{j}}=(36+\lambda)\hat{\text{i}}+(3\mu+3)\hat{\text{j}}$
Equating the corresponding components, we get
$36+\lambda=44$
$\Rightarrow\lambda=44-36=8$
and
$3\mu+3=-12$
$\Rightarrow3\mu=-12-3$
$\Rightarrow\mu=-5$
Thus the values of $\lambda\text{ and }\mu$ are 8 and -5, respectively.
View full question & answer→Question 865 Marks
Using vectors find the area of the triangle with vertices, A(2, 3, 5), B(3, 5, 8) and C(2, 7, 8).
AnswerLet $\vec{\text{a}},\vec{\text{b}},$ and $\vec{\text{c}}$ be the position vectors of A, B and C, respectively. then,
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+5\hat{\text{k}}$
$\vec{\text{b}}=3\hat{\text{i}}+5\hat{\text{j}}+8\hat{\text{k}}$
$\vec{\text{c}}=2\hat{\text{i}}+7\hat{\text{j}}+8\hat{\text{k}}$
Now,
$\overrightarrow{\text{AB}}=\vec{\text{b}}-\vec{\text{a}}$
$=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
$\overrightarrow{\text{AC}}=\vec{\text{c}}=\vec{\text{a}}$
$=0\hat{\text{i}}+4\hat{\text{j}}+3\hat{\text{k}}$
$\therefore\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&2&3\\0&4&3 \end{vmatrix}$
$=-6\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
$\Rightarrow\big|\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}\big|=\sqrt{36+9+16}$
$=\sqrt{61}$
Area of triangle ABC $=\frac{1}{2}\big|\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}\big|$
$=\frac{\sqrt{61}}{2}\text{ sq. units}$
View full question & answer→Question 875 Marks
Show that the vectors
$\vec{\text{a}}=\frac{1}{7}(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}),\vec{\text{b}}=\frac{1}{7}(3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}),\vec{\text{c}}=\frac{1}{7}(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})$ are mutually perpendicular unit vectors.
AnswerWe have
$|\vec{\text{a}}|=\frac{1}{7}\sqrt{2^2+3^2+6^2}=\frac{1}{7}\sqrt{49}=\frac{7}{7}=1$
$\big|\vec{\text{b}}\big|=\frac{1}{7}\sqrt{3^2+(-6)^2+2^2}=\frac{1}{7}\sqrt{49}=\frac{7}{7}=1$
$|\vec{\text{c}}|=\frac{1}{7}\sqrt{6^2+2^2+(-3)^2}=\frac{1}{7}\sqrt{49}=\frac{7}{7}=1$
And
$\vec{\text{a}}.\vec{\text{b}}$
$=\frac{1}{7}(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}).\frac{1}{7}(3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}})$
$=\frac{1}{49}(6-18+12)$
$=0$
$\vec{\text{b}}.\vec{\text{c}}$
$=\frac{1}{7}(3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}).\frac{1}7{}(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})$
$=\frac{1}{49}(18-12-6)$
$=0$
$\vec{\text{c}}.\vec{\text{a}}$
$=\frac{1}{7}(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}).\frac{1}7{}(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}})$
$=\frac{1}{49}(12+6-18)$
$=0$
So, $|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=|\vec{\text{c}}|=1$ and $\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0$
So, the given vectors are mutuaiiy perpendicular unit vectors.
View full question & answer→Question 885 Marks
The two adjacent sides of a parallelogram are $2\hat{\text{i}}-4\hat{\text{j}}-5\hat{\text{k}}$ and $2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}.$ Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram.
AnswerThe two adjacent sides of a parallelogram are $2\hat{\text{i}}-4\hat{\text{j}}-5\hat{\text{k}}$ and $2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}.$ suppose $\vec{\text{a}}=2\hat{\text{i}}-4\hat{\text{j}}-5\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
Then any one diagonal of a parallelogram is $\vec{\text{P}}=\vec{\text{a}}+\vec{\text{b}}.$
$\vec{\text{P}}=\vec{\text{a}}+\vec{\text{b}}$
$=2\hat{\text{i}}-4\hat{\text{j}}-5\hat{\text{k}}+2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
$=4\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$
Therefore, unit vector along the diagonal is $\frac{\vec{\text{P}}}{\big|\vec{\text{P}}\big|}=\frac{4\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}}{\sqrt{16+4+4}}=\frac{2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}}{\sqrt{6}}.$
Another diagonal of a parallelogram is $\vec{\text{P}}'=\vec{\text{b}}-\vec{\text{a}}.$
$\vec{\text{P}}'=\vec{\text{b}}-\vec{\text{a}}$
$=2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}-2\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}$
$=6\hat{\text{j}}+8\hat{\text{k}}$
Therefore, unit vector along the diagonal is $\frac{\vec{\text{P}}}{\big|\vec{\text{P}}\big|}=\frac{6\hat{\text{j}}+8\hat{\text{k}}}{\sqrt{36+64}}=\frac{6\hat{\text{j}}+8\hat{\text{k}}}{10}=\frac{3\hat{\text{j}}+4\hat{\text{k}}}{5}.$
Now,
$\vec{\text{P}}\times\vec{\text{P}}'=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\4&-2&-2\\0&6&8\end{vmatrix}$
$=\hat{\text{i}}(-16+12)-\hat{\text{j}}(32-0)+\hat{\text{k}}(24-0)$
$=-4\hat{\text{i}}-32\hat{\text{j}}+24\hat{\text{k}}$
Area of parallelogram $\frac{\big|\vec{\text{p}}\times\vec{\text{p}'}\big|}{2}=\frac{\sqrt{16+1024+576}}{2}=\frac{\sqrt{1616}}{2}$
$=\frac{4\sqrt{101}}{2}=2\sqrt{101}\text{ square units}$ View full question & answer→Question 895 Marks
If four points A, B, C and D with position vectors $4\hat{\text{i}}+3\hat{\text{j}},5\hat{\text{i}}+\text{x}\hat{\text{j}}+7\hat{\text{k}},5\hat{\text{i}}+3\hat{\text{j}}$ and $7\hat{\text{i}}+6\hat{\text{j}}+\hat{\text{k}}$ respectively are coplanar, then find the value of x.
AnswerLet $\vec{\text{OA}}=4\vec{\text{i}}+3\vec{\text{j}}+3\vec{\text{k}},\vec{\text{OB}}=5\hat{\text{i}}+\text{x}\hat{\text{j}}+7\hat{\text{k}},\vec{\text{OC}}=5\hat{\text{i}}+3\hat{\text{j}}$ and $7\hat{\text{i}}+6\hat{\text{j}}+\hat{\text{k}}.$
$\therefore\vec{\text{AB}}=(5\hat{\text{i}}+\text{x}\hat{\text{j}}+7\hat{\text{k}})-(4\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}})\\=\hat{\text{i}}+(\text{x}-3)\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{AC}}=(5\hat{\text{i}}+3\hat{\text{j}})-(4\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}})\\=\hat{\text{i}}-3\hat{\text{k}}$
$\vec{\text{AD}}=(7\hat{\text{i}}+6\hat{\text{j}}+\hat{\text{k}})-(4\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}})\\=3\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}}$
Since the given four points are coplanar, so the vectors $\vec{\text{AB}},\vec{\text{AC}}$ and $\vec{\text{AD}}$ are also coplanar.
$\therefore\big[\vec{\text{AB}}\vec{\text{ AC }}\vec{\text{AD}}\big]=0$
$\begin{vmatrix}1&\text{x}-3&4\\1&0&-3\\3&3&-2 \end{vmatrix}=0$
$\Rightarrow 1(0+9)-(\text{x}-3)(-2+9)+4(3-0)=0$
$\Rightarrow 9-7\text{x}+21+12=0$
$\Rightarrow7\text{x}=42$
$\Rightarrow \text{x}=6$
Thus, the value of x is 6.
View full question & answer→Question 905 Marks
Show that the vectors $2\hat{i}-\hat{j}+\hat{k},\ \hat{i}-3\hat{j}-5\hat{k}\ \text{and}\ 3\hat{i}-4\hat{j}-4\hat{k}$ form the vertices of a right angled triangle.
AnswerLet $\vec{a}=2\hat{i}-\hat{j}+\hat{k},\ \vec{b}=\hat{i}-3\hat{j}-5\hat{k},\ \vec{c}=3\hat{i}-4\hat{j}-4\hat{k}$be the position vectors of A, B, C respectively.
$\therefore\ \overrightarrow{\text{AB}}=\text{P.V. of B - P.V. of A}=\vec{b}-\vec{a}$$=(\hat{i}-3\hat{j}-5\hat{k})-(2\hat{i}-\hat{j}+\hat{k})=-\hat{i}-2\hat{j}-6\hat{k}$
$\overrightarrow{\text{BC}}=\text{P.V. of C - P.V. of B}=\vec{c}-\vec{b}$$=(3\hat{i}-4\hat{j}-4\hat{k})-(\hat{i}-3\hat{j}-5\hat{k})=2\hat{i}-\hat{j}+\hat{k}$
$\overrightarrow{\text{CA}}=\text{P.V. of A - P.V. of C}=\vec{a}-\vec{c}$ $=(2\hat{i}-\hat{j}+\hat{k})-(3\hat{i}-4\hat{j}-4\hat{k})=-\hat{i}+3\hat{j}+5\hat{k}$
$\therefore\ \ \overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\hat{i}-3\hat{j}-5\hat{k}=-\overrightarrow{\text{CA}}$ $\therefore\ \ \overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{CA}}+\vec{0}$ $\text{Again},\ \overrightarrow{\text{BC}}.\overrightarrow{\text{CA}}=(2)(-1)+(-1)(3)+(1)(5)$ $=-2-3+5=0$ $\therefore\ \ \overrightarrow{\text{BC}}\ \text{and}\ \overrightarrow{\text{CA}}$ are perpendicular $\therefore\ \ \triangle\text{ABC}$ is right angled $\therefore$ given position vectors $\vec{a}, \vec{b}, \vec{c}$ form the vertices of a right angled triangle.
View full question & answer→Question 915 Marks
Given $\vec{\text{a}}=\frac{1}{7}\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big),\vec{\text{b}}=\frac{1}{7}\big(3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}\big),$$\vec{\text{c}}=\frac{1}{7}\big(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big),\hat{\text{i}},\hat{\text{j}},\hat{\text{k}}$
being a right handed orthogonal system of unit vector in spece, show that $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ is also another system.
AnswerGiven:
$\vec{\text{a}}=\frac{1}{7}\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)$
$\vec{\text{b}}=\frac{1}{7}\big(3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}\big)$
$\vec{\text{c}}=\frac{1}{7}\big(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)$
$\vec{\text{a}}\times\vec{\text{b}}=\Big(\frac{1}{7}\Big)\Big(\frac{1}{7}\Big)\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&3&6\\3&-6&2 \end{vmatrix}$
$=\frac{1}{49}\big(42\hat{\text{i}}+14\hat{\text{j}}-21\hat{\text{k}}\big)$
$=\frac{1}{49}\big[7\big(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)\big]$
$=\frac{1}{7}\big(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)$
$=\vec{\text{c}}$
$\vec{\text{b}}\times\vec{\text{c}}=\Big(\frac{1}{7}\Big)\Big(\frac{1}{7}\Big)\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&-6&2\\6&2&-3 \end{vmatrix}$
$=\frac{1}{49}\big(14\hat{\text{i}}+21\hat{\text{j}}+42\hat{\text{k}}\big)$
$=\frac{1}{49}\big[7\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)\big]$
$=\frac{1}{7}\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)$
$=\vec{\text{a}}$
$\vec{\text{c}}\times\vec{\text{a}}=\Big(\frac{1}{7}\Big)\Big(\frac{1}{7}\Big)\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\6&2&-3\\2&3&6 \end{vmatrix}$
$=\frac{1}{49}\big(21\hat{\text{i}}-42\hat{\text{j}}+14\hat{\text{k}}\big)$
$=\frac{1}{49}\big[7\big(3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}\big)\big]$
$=\frac{1}{7}\big(3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}\big)$
$=\vec{\text{b}}$
$|\vec{\text{a}}|=\frac{1}{7}\sqrt{4+9+36}$
$=\frac{7}{7}$
$=1$
$\big|\vec{\text{b}}\big|=\frac{1}{7}\sqrt{9+36+4}$
$=\frac{7}{7}$
$=1$
$|\vec{\text{c}}|=\frac{1}{7}\sqrt{36+4+9}$
$=\frac{7}{7}$
$=1$
Thus, $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ from a right handed orthogonal system of unit vectors.
View full question & answer→Question 925 Marks
Prove that the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.
Answer
Let ABCD be a rectangle.Take A as origion.
Let position vectors of point B, D be $\vec{\text{a}}$ and $\vec{\text{b}}$ respectively.
By parallelogram law,
$\vec{\text{AC}} = \vec{\text{a}} + \vec{\text{b}}$ and $\vec{\text{BD}} = \vec{\text{a}} - \vec{\text{b}}$
As ABCD is a rectangle, $\text {AB}\perp{\text{AD}}$
$\Rightarrow \vec{\text{a}}. \vec{\text{b}} = 0\dots(\text{i})$
Now, diagonals AC and BD are perpendicular iff $\vec{\text{AC}} . \vec{\text{BD}} = 0$
$\Rightarrow \big(\vec{\text{a}} + \vec{\text{b}}\big)\big(\vec{\text{a}} - \vec{\text{b}}\big) = 0$
$\Rightarrow (\vec{\text{a}})^{2} - (\vec{\text{b}})^{2} = 0$
$\Rightarrow\big|\vec{\text{AB}}\big|^{2} = \big|\vec{\text{AD}}\big|^{2}$
$\Rightarrow \big|\vec{\text{AB}}\big|^{2} = \big|\vec{\text{AD}}\big|$
Hence ABCD is a square. View full question & answer→Question 935 Marks
Show that the four points A, B, C and D with the position vectors $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ and $\vec{\text{d}}$ respectively are coplanar if and only if $3\vec{\text{a}}-2\vec{\text{b}}+\vec{\text{c}}-2\vec{\text{d}}=\vec0$.
AnswerNecessary Condition: Firstly, let $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are coplanar vectors. Then, one of them is expressible as a linear combination of the other two.Let $\vec{\text{c}}=\text{x}\vec{\text{a}}+\text{y}\vec{\text{b}}$ for some scalars x, y. Then,
$\Rightarrow\ \text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=\vec0$, where l = x, m = y, n = -1.
Thus, if $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are coplanar vectors, then there exists a scalars l, m, n not all zero simultaneously satisfying $\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=\vec0$ where l, m, n are not all zero simultaneously.
Sufficient Condition: Let $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are three scalars such that there exists scalars l, m, n not all zero simultaneously satisfying $\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=\vec0$. We have to prove that $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are coplanar vectors.
Now,
$\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=\vec0$
$\Rightarrow\text{n}\vec{\text{c}}=-\text{l}\vec{\text{a}}-\text{m}\vec{\text{b}}$
$\Rightarrow\ \vec{\text{c}}=\Big(\frac{-\text{l}}{\text{n}}\Big)\vec{\text{a}}+\Big(\frac{-\text{m}}{\text{n}}\Big)\vec{\text{b}}$
$\vec{\text{c}}$ is a linear combination of $\vec{\text{a}}\text{ and }\vec{\text{b}}$.
$\vec{\text{c}}$ lies in a plane $\vec{\text{a}}\text{ and }\vec{\text{b}}$.
Hence, $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are coplanar vectors.
View full question & answer→Question 945 Marks
Using vectors, find the area of the triangle with vertices:
- A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5)
- A(1, 2, 3), B(2, -1, 4) and C(4,5, -1)
AnswerGiven that
A = (1, 1, 2)
B = (2, 3, 5)
C = (1, 5, 5)
Position vector of $\text{A}=\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
Position vector of $\text{B}=2\hat{\text{i}}+3\hat{\text{j}}+5\hat{\text{k}}$
Position vector of $\text{C}=\hat{\text{i}}+5\hat{\text{j}}+5\hat{\text{k}}$
$\overrightarrow{\text{AB}}=$ Position vector of B-position vector of A
$=2\hat{\text{i}}+3\hat{\text{j}}+5\hat{\text{k}}-\big(\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$
$=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
$\overrightarrow{\text{AC}}-$ Position vector of C-position vector of A
$=\hat{\text{i}}+5\hat{\text{j}}+5\hat{\text{k}}-\big(\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$
$=4\hat{\text{j}}+3\hat{\text{k}}$
Area of triangle $=\frac{1}{2}\big|\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}\big|$
$\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}\begin{bmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&2&3\\0&4&3 \end{bmatrix}$
$=\hat{\text{i}}(6-12)-\hat{\text{j}}(3-0)+\hat{\text{k}}(4-0)$
$=-6\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
$\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}=\sqrt{(-6)^2+(-3)^2+4^2}$
$-\sqrt{36+9+16}$
$=\sqrt{61}$
Area of the teiangle $=\frac{1}{2}\sqrt{61}$ Sq. unit
View full question & answer→Question 955 Marks
(Pythagoras's theorem) Prove by vector method that in a right angleg triang, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Question 965 Marks
Show that the vectors $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ given by $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\ \vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ are non-coplanar. Express vector $\vec{\text{d}}=2\hat{\text{i}}-\hat{\text{j}}-3\hat{\text{k}}$ as a linear combination of the vectors $\vec{\text{a}},\ \vec{\text{b}}\text{ and }\vec{\text{c}}$.
AnswerLet the given vectors $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\ \vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ are coplanar. Then one of the vector is expressible as a linear combination of the other two. Let,
$\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}=\text{x}\big(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big)+\text{y}\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
$=\hat{\text{i}}(2\text{x + y})+\hat{\text{j}}(\text{x + y})+\hat{\text{k}}(3\text{x + y})$
$\Rightarrow2\text{x + y}=1,\ \text{x + y}=2,\ 3\text{x + y}=3$
On solving the first two equations we get x = -1, y = 3. Clearly the values of x and y does not satisfy the third equation.
Hence the given vector are non-coplanar.
Now, $\vec{\text{d}}=2\hat{\text{i}}-\hat{\text{j}}-3\hat{\text{k}}$ which can be expressed as
$2\hat{\text{i}}-\hat{\text{j}}-3\hat{\text{k}}=\text{x}\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)\\+\text{y}\big(2\text{i}+\hat{\text{j}}+3\hat{\text{k}}\big)+\text{z}\big(\text{i}+\hat{\text{j}}+\hat{\text{k}}\big)$
$=\text{i}(\text{x}+2\text{y}+\text{z})+\hat{\text{j}}(2\text{x + y + z})+\hat{\text{k}}(3\text{x}+3\text{y + z})$
$\Rightarrow\ \text{x}+2\text{y}+\text{z}=2,\\2\text{x + y + z}=-1,\ 3\text{x}+3\text{y + z}=-3$
$\Rightarrow\ \text{x}=-\frac{8}3,\ \text{y}=\frac{1}3,\ \text{z}=4$
Hence $\vec{\text{d}}$ is expressible as the linear combination of $\vec{\text{a}},\ \vec{\text{b}}\text{ and }\vec{\text{c}}$.
View full question & answer→Question 975 Marks
In a quadrilateral $ABCD$, prove that $AB^2+ BC^2+ CD^2+ DA^2= AC^2+ BD^2+ 4PQ^2$, where $P$ and $Q$ are middle points of diagonals $AC$ and $BD$.
Answer
Take O as origin, let the position vectors of A, B C and D are $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ and $\vec{\text{d}}$ respectively.position vector of $\text{p} = \frac{\vec{\text{a}}+ \vec{\text{c}}}{2}$position vector of $\text{Q} = \frac{\vec{\text{a}}+\vec{\text{d}}}{2}$
$LHS = AB^2 + BC^2 + CD^2 + DA^2$
$ = (\vec{\text{b}} - \vec{\text{a}})^{2} + (\vec{\text{c}}-\vec{\text{b}})^{2}+(\vec{\text{d}}-\vec{\text{c}})^{2}+(\vec{\text{d}}-\vec{\text{a}})^{2}$
$= 2\big[(\vec{\text{a}})^{2}+(\vec{\text{b}})^{2}+(\vec{\text{c}})^{2}+(\vec{\text{d}})^{2}-\vec{\text{a}}\vec{\text{b}}\cos\theta_1-\vec{\text{b}}\vec{\text{c}}\cos\theta_2-\vec{\text{d}}\vec{\text{c}}\cos\theta_1-\vec{\text{c}}\vec{{\text{a}}}\cos\theta_4\big]$
$RHS = AC^2 + BD^2 + 4PQ^2$
$ = (\vec{\text{c}}-\vec{\text{a}})^{2} + (\vec{\text{d}}-\vec{\text{b}})^{2} + 4\Big(\frac{\vec{\text{a}} + \vec{\text{b}}}{2}-\frac{\vec{\text{a}}+\vec{\text{c}}}{2}\Big)^{2}$
$= 2 \big[(\vec{\text{a}})^{2} +(\vec{\text{b}})^{2} + (\vec{\text{c}})^{2}+ (\vec{\text{d}})^{2}-\vec{\text{a}}\vec{\text{b}}\cos\theta_1-\vec{\text{b}}\vec{\text{c}}\cos\theta_2-\vec{\text{d}}\vec{\text{c}}\cos\theta_1-\vec{\text{c}}\vec{\text{a}}\cos\theta_4\big]$
= LHS
Hence proved. View full question & answer→Question 985 Marks
Show that the four points P, Q, R, S with position vectors $\vec{\text{p}},\ \vec{\text{q}},\ \vec{\text{r}},\ \vec{\text{s}}$ respectively such that $5\vec{\text{p}}-2\vec{\text{q}}+6\vec{\text{r}}-9\vec{\text{s}}=0$, are coplanar. Also, find the position vector of the point of intersection of the line segments PR and QS.
AnswerWe have given that,
$5\vec{\text{p}}-2\vec{\text{q}}+6\vec{\text{r}}-9\vec{\text{s}}=0$
Where $\vec{\text{p}},\ \vec{\text{q}},\ \vec{\text{r}}\text{ and }\vec{\text{s}}$ are the position vectors of point P, Q, R and S.
$5\vec{\text{p}}+6\vec{\text{r}}=2\vec{\text{q}}+9\vec{\text{s}}\ \dots(\text{i})$
Sum of the co-efficients on both sides of the equation (i) is 11. So. divide equation (i) by 11 on both the sides.
$\frac{5\vec{\text{p}}+6\vec{\text{r}}}{11}=\frac{2\vec{\text{q}}+9\vec{\text{s}}}{11}$
$\frac{5\vec{\text{p}}+6\vec{\text{r}}}{5+6}=\frac{2\vec{\text{q}}+9\vec{\text{s}}}{2+9}$
It shows that position vector of a point A dividing PR in the ratio of 6 : 5 and QS in the ratio of 9 : 2. Thus, A is the common point to PR and QS and it is also point of intersection of PQ and QS.
So,
P, Q, R and S are coplanar.
Position vector of point A is given by
$\frac{5\text{p}+6\text{q}}{11}\text{ or }\frac{2\vec{\text{q}}+9\vec{\text{s}}}{11}$
View full question & answer→Question 995 Marks
If O is a point in space, ABC is a triangle and D, E, F are the mid-points of the sides BC, CA and AB respectively of the triangle, prove that $\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}=\overrightarrow{\text{OD}}+\overrightarrow{\text{OE}}+\overrightarrow{\text{OF}}$.
Answer
Let D, E and F are the midpoints of BC, CA and AB respectively.
Therefore,
$\frac{\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}}{2}=\overrightarrow{\text{OD}}$
$\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}=2\ \overrightarrow{\text{OD}}\ \dots(1)$
Similarly
$\overrightarrow{\text{OC}}+\overrightarrow{\text{OA}}=2\ \overrightarrow{\text{OE}}\ \dots(2)$
$\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}=2\ \overrightarrow{\text{OF}}\ \dots(3)$
Adding (1), (2) and (3). We get,
$2\Big(\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}\Big)=2\Big(\overrightarrow{\text{OD}}+\overrightarrow{\text{OE}}+\overrightarrow{\text{OF}}\Big)$
$\Rightarrow\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}=\overrightarrow{\text{OD}}+\overrightarrow{\text{OE}}+\overrightarrow{\text{OF}}$
Hence Proved. View full question & answer→Question 1005 Marks
If AD is the median of $\triangle\text{ABC},$ using vectors, prove that $\text{AB}^2+\text{AC}^2=2\big(\text{AD}^2+\text{CD}^2\big).$
Answer
Take A as origin, let the position vectors of B and C are $\vec{\text{b}}$ and $\vec{\text{c}}$ respectively. position vector of $\text{D} = \frac{\vec{\text{b}} + \vec{\text{c}}}{2},\vec{\text{AB}} = \vec{\text{b}}$ and $\vec{\text{AC}} = \vec{\text{c}.}$$\vec{\text{AD}} = \frac{\vec{\text{b}} + \vec{\text{c}}}{2} - 0 = \frac{\vec{\text{b}} +\vec{\text{c}}}{2}$
consider, $2 (AD^2 + CD^2)$
$= 2\Big[\Big(\frac{\vec{\text{b}}+ \vec{\text{c}}}{2}\Big)^{2} + \Big(\frac{\vec{\text{b}} + \vec{\text{c}}}{2}-\vec{\text{c}}\Big)^{2}\Big]$
$=2\Big[\Big(\frac{\vec{\text{b}} + \vec{\text{c}}}{2}\Big)^{2}\Big(\frac{\vec{\text{b}} - \vec{\text{c}}}{2}\Big)^{2}\Big]$
$=\frac{1}{2}\big[\big(\vec{\text{b}} + \vec{\text{c}}\big)^{2} + \big(\vec{\text{b}} - \vec{\text{c}}\big)^{2}\big]$
$= (\vec{\text{b}})^{2} + (\vec{\text{c}})^{2}$
$= \big(\vec{\text{AB}}\big)^{2} + \big(\vec{\text{AC}}\big)^{2}$
$= \text{AB}^2 + \text{AC}^2$
Hence proved. View full question & answer→