Question 15 Marks
Prove that the points $\hat{\text{i}}-\hat{\text{j}},\ 4\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$ and $2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}$ are the vertices of a right-angled triangle.
AnswerLet $\vec{\text{A}}=\hat{\text{i}}-\hat{\text{j}}$
$\vec{\text{B}}=4\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{C}}=2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}$
$\overrightarrow{\text{AB}}=\vec{\text{B}}-\vec{\text{A}}$
$=\big(4\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}\big)-\big(\hat{\text{i}}-\hat{\text{j}}\big)$
$=4\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}-\hat{\text{i}}-\hat{\text{j}}$
$=3\hat{\text{i}}+4\hat{\text{j}}+\hat{\text{k}}$
$\Big|\overrightarrow{\text{AB}}\Big|=\sqrt{(3^2)+(4)^2+(1)^2}$
$=\sqrt{9+16+1}$
$=\sqrt{26}$
$\overrightarrow{\text{BC}}=\vec{\text{C}}-\vec{\text{B}}$
$=\big(\hat{2\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}\big)-\big(4\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}\big)$
$=2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}-4\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}}$
$=-2\hat{\text{i}}-7\hat{\text{j}}+4\hat{\text{k}}$
$\Big|\overrightarrow{\text{BC}}\Big|=\sqrt{(2)^2+(-7)^2+(4)^2}$
$=\sqrt{4+49+16}$
$=\sqrt{69}$
$\overrightarrow{\text{CA}}=\vec{\text{A}}-\vec{\text{C}}$
$=\hat{\text{i}}-\hat{\text{j}}-\big(2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}\big)$
$=\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}$
$=-\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}$
$\Big|\overrightarrow{\text{CA}}\Big|=\sqrt{(-1)^2+(3)^2+(-5)^2}$
$=\sqrt{1+9+25}$
$=\sqrt{35}$
Here, $\Big|\overrightarrow{\text{AB}}\Big|^2+\Big|\overrightarrow{\text{CA}}\Big|^2=\Big|\overrightarrow{\text{BC}}\Big|^2$
$26+35=69$
$61\neq69$
$\text{LHS}\neq\text{RHS}$
Since sum of square of two sides is not equal to the square of third sides. So, $\triangle\text{ABC}$ is not a right triangle.
View full question & answer→Question 25 Marks
The adjacent sides of a parallelogram are represented by the vectors $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{b}}=-2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$. Find the unit vectors parallel to the diagonals of the parallelogram.
AnswerLet PQRS be a parallelogram such that $\text{PQ}=\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$and $\text{QR}=\vec{\text{b}}=-2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
In $\triangle\text{PQR}$ $\overrightarrow{\text{PQ}}+\overrightarrow{\text{QR}}=\overrightarrow{\text{PR}}$ $\overrightarrow{\text{PR}}=\vec{\text{a}}+\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}+\big(-2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$ $\overrightarrow{\text{PR}}=-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$ In $\triangle{\text{PQS}}$ $\overrightarrow{\text{PS}}+\overrightarrow{\text{SQ}}=\overrightarrow{\text{PQ}}$ $\overrightarrow{\text{SQ}}=\vec{\text{a}}-\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}-\big(-2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$ $\overrightarrow{\text{SQ}}=3\hat{\text{i}}+0\hat{\text{j}}-3\hat{\text{k}}$ The unit vector along $\overrightarrow{\text{PR}}=\frac{\overrightarrow{\text{PR}}}{\big|\overrightarrow{\text{PR}}\big|}=\frac{-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}}{\sqrt{1+4+1}}$ $=\frac{1}{\sqrt6}\big(-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)$ The unit vector along $\overrightarrow{\text{SQ}}=\frac{\overrightarrow{\text{SQ}}}{\big|\overrightarrow{\text{SQ}}\big|}=\frac{3\hat{\text{i}}+0\hat{\text{j}}-3\hat{\text{k}}}{\sqrt{9+0+9}}$ $=\frac{1}{\sqrt2}\big(\hat{\text{i}}-\hat{\text{k}}\big)$ View full question & answer→Question 35 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are non-zero, non-coplanar vectors, prove that the vector is coplanar:
$5\vec{\text{a}}+6\vec{\text{b}}+7\vec{\text{c}},\ 7\vec{\text{a}}-8\vec{\text{b}}+9\vec{\text{c}}$ and $3\vec{\text{a}}+20\vec{\text{b}}+5\vec{\text{c}}$
AnswerWe know that, Three vectors are coplanar if one of the vector can be expressed as the linear combination of other two. Let, $5\vec{\text{a}}+6\vec{\text{b}}+7\vec{\text{c}}\\=\text{x}\big(7\vec{\text{a}}-8\vec{\text{b}}+9\vec{\text{c}}\big)+\text{y}\big(3\vec{\text{a}}+20\vec{\text{b}}+5\vec{\text{c}}\big)$ $5\vec{\text{a}}+6\vec{\text{b}}+7\vec{\text{c}}\\=7\vec{\text{a}}\text{x}-8\vec{\text{b}}\text{x}+9\vec{\text{c}}\text{x}+3\vec{\text{a}}\text{y}+20\vec{\text{b}}\text{y}+5\vec{\text{c}}\text{y}$ $5\vec{\text{a}}+6\vec{\text{b}}+7\vec{\text{c}}\\=\big(7\text{x}+3\text{y}\big)\vec{\text{a}}+\big(-8\text{x}+20\text{y}\big)\vec{\text{b}}+\big(9\text{x}+5\text{y}\big)\vec{\text{c}}$ Comparing the LHS and RHS, 7x + 3y = 5 .....(i) -8x + 20y = 6 .....(ii) 9x + 5y = 7 .....(iii) For solving (i) and (ii), Subtract -8 × (i) from 7 × (ii),
$\text{y}=\frac{82}{164}$ $\text{y}=\frac{1}2$ Put $\text{y}=\frac{1}2$ in equation (i), $7\text{x}+3\text{y}=5$ $7\text{x}+3\Big(\frac{1}2\Big)=5$ $7\text{x}+\frac{3}2=5$ $7\text{x}=\frac{5}1-\frac{3}2$ $7\text{x}=\frac{10-3}2$ $7\text{x}=\frac{7}2$ $\text{x}=\frac{7}{14}$ $\text{x}=\frac{1}{2}$ Now, put $\text{x}=\frac{1}{2}$ and $\text{y}=\frac{1}2$ in equation (iii), $9\text{x}+5\text{y}=7$ $9\Big(\frac{1}2\Big)+5\Big(\frac{1}2\Big)=7$ $\frac{9}2+\frac{5}2=7$ $\frac{14}2=7$ $7=7$LHS = RHS
$\therefore$ The value of x, y satisfy equation (iii).
So, $5\vec{\text{a}}+6\vec{\text{b}}+7\vec{\text{c}},\ 7\vec{\text{a}}-8\vec{\text{b}}+9\vec{\text{c}},\ 3\vec{\text{a}}+20\vec{\text{b}}+5\vec{\text{c}}$ are coplanar. View full question & answer→Question 45 Marks
If $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are non-coplanar vectors, prove that the point having the following position vectors is collinear:$\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}},\ 4\vec{\text{a}}+3\vec{\text{b}},\ 10\vec{\text{a}}+7\vec{\text{b}}-2\vec{\text{c}}$
AnswerLet the points be A, B, C
Position vector of $\text{A}=\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$
Position vector of $\text{B}=4\vec{\text{a}}+3\vec{\text{b}}$
Position vector of $\text{C}=10\vec{\text{a}}+7\vec{\text{b}}-2\vec{\text{c}}$
$\overrightarrow{\text{AB}}=$Position vector of B - Postion vector of A
$=\big(4\vec{\text{a}}+3\vec{\text{b}}\big)-\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)$
$=4\vec{\text{a}}+3\vec{\text{b}}-\vec{\text{a}}-\vec{\text{b}}-\vec{\text{c}}$
$\overrightarrow{\text{AB}}=3\vec{\text{a}}+2\vec{\text{b}}-\vec{\text{c}}$
$\overrightarrow{\text{BC}}=$Position vector of C - Postion vector of B
$=\big(10\vec{\text{a}}+7\vec{\text{b}}-2\vec{\text{c}}\big)-\big(4\vec{\text{a}}+3\vec{\text{b}}\big)$
$=10\vec{\text{a}}+7\vec{\text{b}}-2\vec{\text{c}}-4\vec{\text{a}}-3\vec{\text{b}}$
$\overrightarrow{\text{BC}}=6\vec{\text{a}}+4\vec{\text{b}}-2\vec{\text{c}}$
Using $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$
$\overrightarrow{\text{BC}}=2\Big(\overrightarrow{\text{AB}}\Big)$
So, $\overrightarrow{\text{AB}}$ is parallel to $\overrightarrow{\text{BC}}$ but $\vec{\text{B}}$ is a common vector. Hence,
A, B, C are collinear.
View full question & answer→Question 55 Marks
If a unit vector $\vec{\text{a}}$ makes an angle $\frac{\pi}3$ with $\hat{\text{i}}$, $\frac{\pi}4$ with $\hat{\text{j}}$ and an acute angle $\theta$ with $\hat{\text{k}}$, then find $\theta$ and hence, the components of $\vec{\text{a}}$.
AnswerLet unit vector $\vec{\text{a}}$ have $\left(a_1, a_2, a_3\right)$ components.
$\Rightarrow\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
Since $\vec{\text{a}}$ is a unit vector, $|\vec{\text{a}}|=1$
Also it is given that $\vec{\text{a}}$ makes angles $\frac{\pi}3$ with $\hat{\text{i}}$, $\frac{\pi}4$ with $\hat{\text{j}}$, and an acute angle $\theta$ with $\hat{\text{k}}$.
Then, we have:
$\cos\frac{\pi}3=\frac{\text{a}_1}{|\vec{\text{a}}|}$
$\Rightarrow\ \frac{1}2=\text{a}_1$ $[|\vec{\text{a}}|=1]$
$\cos\frac{\pi}4=\frac{\text{a}_2}{|\vec{\text{a}}|}$
$\Rightarrow\ \frac{1}{\sqrt2}=\text{a}_2$ $[|\vec{\text{a}}|=1]$
Also, $\cos\theta=\frac{\text{a}_3}{|\vec{\text{a}}|}$
$\Rightarrow\ \text{a}_3=\cos\theta$
Now,
$|\text{a}|=1$
$\Rightarrow\ \sqrt{\text{a}_1^2+\text{a}_2^2+\text{a}_3^2}=1$
$\Rightarrow\Big(\frac{1}2\Big)^2+\Big(\frac{1}{\sqrt{2}}\Big)^2+\cos^2\theta=1$
$\Rightarrow\ \frac{1}4+\frac{1}2+\cos^2\theta=1$
$\Rightarrow\ \frac{3}4+\cos^2\theta=1$
$\Rightarrow\ \cos^2\theta=1-\frac{3}4=\frac{1}4$
$\Rightarrow\ \cos\theta=\frac{1}2$
$\Rightarrow\ \theta=\frac{\pi}3$
$\therefore\ \text{a}_3=\cos\frac{\pi}3=\frac{1}2$
Hence, $\theta=\frac{\pi}3$ and the components of $\vec{\text{a}}$ are $\Big(\frac{1}2,\frac{1}{\sqrt2},\frac{1}{\sqrt2}\Big)$.
View full question & answer→Question 65 Marks
Five forces $\overrightarrow{\text{AB}},\ \overrightarrow{\text{AC}},\ \overrightarrow{\text{AD}},\ \overrightarrow{\text{AE}}\text{ and }\overrightarrow{\text{AF}}$ act at the vertex of a regular hexagon ABCDEF. Prove that the resultant is:$6\ \overrightarrow{\text{AO}}$, where o is the center of hexagon.
Answer
$\overrightarrow{\text{AB}}+\overrightarrow{\text{AC}}+\overrightarrow{\text{AD}}+\overrightarrow{\text{AE}}+\overrightarrow{\text{AF}}$
Consider $\triangle\text{ADE}$,
$\overrightarrow{\text{AD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}=0$
$\overrightarrow{\text{AD}}+\overrightarrow{\text{DE}}=\overrightarrow{\text{AE}}$
$2\ \overrightarrow{\text{AO}}+\Big(-\overrightarrow{\text{AB}}\Big)=\overrightarrow{\text{AE}}$ $\Big[\overrightarrow{\text{AD}}=2\ \overrightarrow{\text{AO}}\text{ and }\overrightarrow{\text{ED}}\ \Big|\Big|\ \overrightarrow{\text{AB}}+\overrightarrow{\text{DE}}=-\overrightarrow{\text{AB}}\Big]$
$\therefore\ \overrightarrow{\text{AE}}+\overrightarrow{\text{AB}}=2\ \overrightarrow{\text{AO}}\ \dots(1)$
Now, consider $\triangle\text{ADC}$,
$\overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DA}}=0$
$\overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}=\overrightarrow{\text{AD}}$ $\Big[\because\ \overrightarrow{\text{CD}}=\overrightarrow{\text{AF}}\Big]$
$\overrightarrow{\text{AC}}+\overrightarrow{\text{AF}}=2\ \overrightarrow{\text{AO}}\ \dots(2)$
Using (1) and (2),
$\overrightarrow{\text{AB}}+\overrightarrow{\text{AE}}+\overrightarrow{\text{AC}}+\overrightarrow{\text{AF}}+\overrightarrow{\text{AD}}$
$2\ \overrightarrow{\text{AO}}+2\ \overrightarrow{\text{AO}}+2\ \overrightarrow{\text{AO}}$
$=6\ \overrightarrow{\text{AO}}$ View full question & answer→Question 75 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 85 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 95 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 105 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 115 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 125 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 135 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 145 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 155 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 165 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 175 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 185 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 195 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 205 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 215 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 225 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 235 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 245 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 255 Marks
Show that the points whose position vectors are as given below are collinear:
$2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\ 3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ and $\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$
AnswerLet the points be A, B and C with position vectors $2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\ 3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ and $\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$. Then, $\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A$=3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}-2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$=\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$
$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}-3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
$=-2\hat{\text{i}}+6\hat{\text{j}}-4\hat{\text{k}}$
$=-2\big(\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}\big)$$\therefore\ \overrightarrow{\text{AB}}=-2\overrightarrow{\text{BC}}$
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are parallel vectors. But B is a point common to them. Hence, A, B, and C are collinear.
View full question & answer→Question 265 Marks
Show that the sum of three vectors determined by the medians of a triangle directed from the vertices is zero.
AnswerLet $\vec{\text{a}}, \vec{\text{b}}\text{ and }\vec{\text{c}}$ are the position vectors of the vertices A, B and C respectively.Then we know that the position vector of the centroid O of the triangle is $\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}{3}$.
Therefore sum of the three vectors $\overrightarrow{\text{OA}},\ \overrightarrow{\text{OB}}\text{ and }\overrightarrow{\text{OC}}$ is $\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}=\vec{\text{a}}-\bigg(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}{3}\bigg)+\vec{\text{b}}-\bigg(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}{3}\bigg)+\vec{\text{c}}-\bigg(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}{3}\bigg)$ $=\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)-3\Big(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3\Big)$$=\vec0$
Hence, Sum of the three vectors determined by the medians of a triangle directed from the vertices is zero.
View full question & answer→Question 275 Marks
Show that the vectors $2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$ and $-4\hat{\text{i}}+6\hat{\text{j}}-8\hat{\text{k}}$ are collinear.
AnswerGiven the position vectors $2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$ and $-4\hat{\text{i}}+6\hat{\text{j}}-8\hat{\text{k}}$Let $\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{b}}=-4\hat{\text{i}}+6\hat{\text{j}}-8\hat{\text{k}}$
Then,
$\vec{\text{b}}=-4\hat{\text{i}}+6\hat{\text{j}}-8\hat{\text{k}}$
$=-2\big(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)$
$=-2\vec{\text{a}}$
Hence, $\vec{\text{a}},\vec{\text{b}}$ are collinear.
View full question & answer→Question 285 Marks
Prove by vector method that the internal bisectors of the angles of a triangle are concurrent.
Answer
Let ABC be a triangle and $\vec\alpha,\ \vec\beta,\ \vec\gamma$ be the position vectors of the vertices A, B and C respectively. Let AD, BE and CF be the internal bisectors of $\angle\text{A},\angle\text{B}\text{ and } \angle\text{C}$ respectively.
We know that D divides BC in the ratio of AB : AC that is c : b.
Then,
P.V. of D is $\frac{\text{c}\vec\gamma+\text{b}\vec\beta}{\text{c}+\text{b}}$.
P.V. of E is $\frac{\text{c}\vec\gamma+\text{a}\vec\alpha}{\text{c}+\text{a}}$.
and P.V. of F is $\frac{\text{a}\vec\alpha+\text{b}\vec\beta}{\text{a}+\text{b}}$.
The point dividing AD in the ratio b + c : a is $\frac{\text{a}\vec\alpha+\text{b}\vec\beta+\text{c}\vec\gamma}{\text{a}+\text{b}+\text{c}}$.
The point dividing BE in the ratio of a + c : b is $\frac{\text{a}\vec\alpha+\text{b}\vec\beta+\text{c}\vec\gamma}{\text{a}+\text{b}+\text{c}}$.
The point dividing CF in the ratio of a + b : c is $\frac{\text{a}\vec\alpha+\text{b}\vec\beta+\text{c}\vec\gamma}{\text{a}+\text{b}+\text{c}}$.
Since the point $\frac{\text{a}\vec\alpha+\text{b}\vec\beta+\text{c}\vec\gamma}{\text{a}+\text{b}+\text{c}}$ lies on all the three internal bisectors AD, BE and CF.
Hence the internal bisectors are concurrent. View full question & answer→Question 295 Marks
Show that the points (3, 4), (-5, 16) and (5, 1) are collinear.
AnswerHere, let A = (3, 4)
B = (-5, 16)
C = (5, 1)
$\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A
$=\big(-5\hat{\text{i}}+16\hat{\text{j}}\big)-\big(3\hat{\text{i}}+4\hat{\text{j}}\big)$
$=-5\hat{\text{i}}+16\hat{\text{j}}-3\hat{\text{i}}-4\hat{\text{j}}$
$\overrightarrow{\text{AB}}=-8\hat{\text{i}}+12\hat{\text{j}}$
$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=\big(5\hat{\text{i}}+\hat{\text{j}}\big)-\big(-5\hat{\text{i}}+16\hat{\text{j}}\big)$
$=5\hat{\text{i}}+\hat{\text{j}}+5\hat{\text{i}}-16\hat{\text{j}}$
$\overrightarrow{\text{BC}}=10\hat{\text{i}}-15\hat{\text{j}} $
So, $4\Big(\overrightarrow{\text{AB}}\Big)=-5\Big(\overrightarrow{\text{BC}}\Big)$
$\overrightarrow{\text{AB}}$ is parallel to $\overrightarrow{\text{BC}}$ but B is a common point.
Hence, A, B, C are collinear.
View full question & answer→Question 305 Marks
In Figure ABCD is a regular hexagon, which vectors are:
- Collinear.
- Equal.
- Co-initial.
- Collinear but not equal.
Answer
- Vectors having the same or parallel supports are called collinear vector. In the given figure the collinear vectors are,
$\vec{\text{a}},\ \vec{\text{d}};\ \vec{\text{x}},\ \vec{\text{z}},\ \vec{\text{b}};\ \vec{\text{c}},\ \vec{\text{y}}$
- vectors having the same magnitude and direction are called equal vector. In the given figure the equal vectors are,
$\vec{\text{b}},\ \vec{\text{x}};\ \vec{\text{c}},\ \vec{\text{y}};\ \vec{\text{a}},\ \vec{\text{d}}$
- Vectors having the same initial point are called co-initial vector. In the given figure the co-initial vectors are,
$\vec{\text{a}},\ \vec{\text{y}},\ \vec{\text{z}}$
- The vectors which are collinear but not equal are $\vec{\text{b}},\ \vec{\text{z}};\ \vec{\text{x}},\ \vec{\text{z}}$
View full question & answer→Question 315 Marks
If the vectors $\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}$ and $\vec{\text{b}}=-6\hat{\text{i}}+\text{m}\hat{\text{j}}$ are collinear, find tghe value of m.
AnswerHere, it is given that vectors $\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}$ and $\vec{\text{b}}=-6\hat{\text{i}}+\text{m}\hat{\text{j}}$ are collinear. So, $\text{a}=\lambda\text{b}$, for a scalar $\lambda$$2\hat{\text{i}}-3\hat{\text{j}}=\lambda\big(-6\hat{\text{i}}+\text{m}\hat{\text{j}}\big)$
$2\hat{\text{i}}-3\hat{\text{j}}=-6\lambda\hat{\text{i}}+\text{m}\lambda\hat{\text{j}}\big)$ Comparing the coefficients of LHS and RHS, $2=-6\lambda$ $\lambda=\frac{2}{-6}$ $\lambda=\frac{-1}3\ \dots(\text{i})$ $-3=\lambda\text{m}$ $\lambda=\frac{-3}{\text{m}}\ \dots(\text{ii})$ From (i) and (ii), $\frac{-1}3=\frac{-3}{\text{m}}$ $\text{m}=3\times3$ $=9$ $\therefore\ \text{m}=9$
View full question & answer→Question 325 Marks
Using vector method, prove that the point is collinear:
A(-3, -2, -5), B(1, 2, 3) and C(3, 4, 7)
AnswerGiven the points A(-3, -2, -5), B(1, 2, 3) and C(3, 4, 7). Then, $\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A$=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}+3\hat{\text{i}}+2\hat{\text{j}}+5\hat{\text{k}}$
$=4\hat{\text{i}}+4\hat{\text{j}}+8\hat{\text{k}}$
$=2\big(2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}\big)$$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=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}}$
$\therefore\ \overrightarrow{\text{AB}}=2\overrightarrow{\text{BC}}$
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are parallel vectors. But B is a point common to them. Hence, the given points A, B, and C are collinear.
View full question & answer→Question 335 Marks
If the points with position vectors $10\hat{\text{i}}+3\hat{\text{j}},\ 12\hat{\text{i}}-5\hat{\text{j}}$ and $\text{a}\hat{\text{i}}+11\hat{\text{j}}$ are collinear, find the value of a.
AnswerLet A, B, C be the points with position vectors $10\hat{\text{i}}+3\hat{\text{j}},\ 12\hat{\text{i}}-5\hat{\text{j}}$ and $\text{a}\hat{\text{i}}+11\hat{\text{j}}$. Then, $\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A $=12\hat{\text{i}}-5\hat{\text{j}}-10\hat{\text{i}}-3\hat{\text{j}}$ $=2\hat{\text{i}}-8\hat{\text{j}}$ $\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B $=\text{a}\hat{\text{i}}+11\hat{\text{j}}-12\hat{\text{i}}+5\hat{\text{j}}$ $=(\text{a}-12)\hat{\text{i}}+16\hat{\text{j}}$ Since, A, B, and C are collinear. $\therefore\ \overrightarrow{\text{AB}}=\lambda\overrightarrow{\text{BC}}$ $\Rightarrow2\hat{\text{i}}-8\hat{\text{j}}=\lambda(\text{a}-12)\hat{\text{i}}+\lambda16\hat{\text{j}}$ $\Rightarrow2=\lambda(\text{a}-12),\ -8=\lambda16$ $\Rightarrow2=\lambda(\text{a}-12),\ \lambda=-\frac{1}2$ $\Rightarrow2=-\frac{1}2(\text{a}-12)$ $\Rightarrow-\text{a}+12=4$ $\Rightarrow\text{a}=8$
View full question & answer→Question 345 Marks
Prove that the given vectors are non-coplanar:
$\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\ 2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}$ and $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
AnswerWe know that, Three vectors are coplanar if any one of them vector can be expressed as the 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)$ $=2\text{x}\hat{\text{i}}+\text{x}\hat{\text{j}}+3\text{x}\hat{\text{k}}+\text{y}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{y}\hat{\text{k}}$ $\therefore\ \hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)\\=\big(2\text{x}+\text{y}\big)\hat{\text{i}}+\big(\text{x}+2\text{y}\big)\hat{\text{j}}+\big(3\text{x}+\text{y}\big)\hat{\text{k}}$ comparing the coefficient of LHS and RHS, 2x + y = 1 .....(i) x + 2y = 2 .....(ii) 3x + y = 3 .....(iii) subtracting 2 × (ii) from equation (i),
$\text{y}=\frac{3}3$ $\text{y}=1$ Put the value of y in equation (i), $2\text{x}+\text{y}=1$ $2\text{x}+1=1$ $2\text{x}=1-1$ $2\text{x}=0$ $\text{x}=\frac{0}2$ $\text{x}=0$ Put the value of x and y in equation (iii), $3\text{x}+\text{y}=3$ $3(0)+1=3$ $0+1=3$ $1=3$ $\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 355 Marks
If the points A(m, -1), B(2, 1) and C(4, 5) are collinear, find the value of m.
AnswerHere, A = (m, -1)
B = (2, 1)
C = (4, 5)
$\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A
$=\big(2\hat{\text{i}}+\hat{\text{j}}\big)-\big(\text{m}\hat{\text{i}}-\hat{\text{j}}\big)$
$=2\hat{\text{i}}+\hat{\text{j}}-\text{m}\hat{\text{i}}+\hat{\text{j}}$
$=(2-\text{m})\hat{\text{i}}+2\hat{\text{j}}$
$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=\big(4\hat{\text{i}}+5\hat{\text{j}}\big)-\big(2\hat{\text{i}}+\hat{\text{j}}\big)$
$=4\hat{\text{i}}+5\hat{\text{j}}-2\hat{\text{i}}-\hat{\text{j}}$
$\overrightarrow{\text{BC}}=2\hat{\text{i}}+4\hat{\text{j}} $
A, B, C are collinear. So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are collinear.
So, $\overrightarrow{\text{AB}}=\lambda\Big(\overrightarrow{\text{BC}}\Big)$
$\big(2-\text{m})\hat{\text{i}}+2\hat{\text{j}}=\lambda\big(2\hat{\text{i}}+4\hat{\text{j}}\big)$, for $\lambda$ scalar
$\big(2-\text{m})\hat{\text{i}}+2\hat{\text{j}}=2\lambda\hat{\text{i}}+4\lambda\hat{\text{j}}$
Comparing the coefficient of LHS and RHS.
$2-\text{m}=2\lambda$
$\frac{2-\text{m}}2=\lambda\ \dots(\text{i})$
$2=4\lambda$
$\frac{2}4=\lambda$
$\frac{1}2=\lambda\ \dots(\text{ii})$
Using (i) and (ii)
$\frac{2-\text{m}}2=\frac{1}2$
$4-2\text{m}=2$
$-2\text{m}=2$
$-2\text{m}=2-4$
$-2\text{m}=-2$
$\text{m}=\frac{-2}{-2}$
$\text{m}=1$
View full question & answer→Question 365 Marks
Show that the points A, B, C with position vectors $\vec{\text{a}}-2\vec{\text{b}}+3\vec{\text{c}},\ 2\vec{\text{a}}+3\vec{\text{b}}-4\vec{\text{c}}$ and $-7\vec{\text{b}}+10\vec{\text{c}}$ are collinear.
AnswerWe have, A, B, C with position vectors $\vec{\text{a}}-2\vec{\text{b}}+3\vec{\text{c}},\ 2\vec{\text{a}}+3\vec{\text{b}}-4\vec{\text{c}}$ and $-7\vec{\text{b}}+10\vec{\text{c}}$ Then,
$\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A
$=2\vec{\text{a}}+3\vec{\text{b}}-4\vec{\text{c}}-\vec{\text{a}}+2\vec{\text{b}}-3\vec{\text{c}}$
$=\vec{\text{a}}+5\vec{\text{b}}-7\vec{\text{c}}$
$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=-7\vec{\text{b}}+10\vec{\text{c}}-2\vec{\text{a}}-3\vec{\text{b}}+4\vec{\text{c}}$
$=-2\vec{\text{a}}-10\vec{\text{b}}+14\vec{\text{c}}$
$=-2\big(\vec{\text{a}}+5\vec{\text{b}}-7\vec{\text{c}}\big)$
$\therefore\ \overrightarrow{\text{BC}}=-2\overrightarrow{\text{AB}}$
Hence, $\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, points A, B and C are collinear.
View full question & answer→Question 375 Marks
If $\vec{\text{a}},\vec{\text{b}}$ are two non-collinear vectors, prove that the points with position vectors $\vec{\text{a}}+\vec{\text{b}},\ \vec{\text{a}}-\vec{\text{b}}$ and $\vec{\text{a}}+\lambda\vec{\text{b}}$ are collinear for all real values of $\lambda$.
AnswerLet A, B, C be the points then, Position vector of $\text{A}=\vec{\text{a}}+\vec{\text{b}}$ Position vector of $\text{B}=\vec{\text{a}}-\vec{\text{b}}$ Position vector of $\text{C}=\vec{\text{a}}+\lambda\vec{\text{b}}$ $\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A $=\big(\vec{\text{a}}-\vec{\text{b}}\big)-\big(\vec{\text{a}}+\vec{\text{b}}\big)$ $=\vec{\text{a}}-\vec{\text{b}}-\vec{\text{a}}-\vec{\text{b}}$ $\overrightarrow{\text{AB}}=-2\vec{\text{b}}$ $\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B $=\big(\vec{\text{a}}+\lambda\vec{\text{b}}\big)-\big(\vec{\text{a}}-\vec{\text{b}}\big)$ $=\vec{\text{a}}+\lambda\vec{\text{b}}-\vec{\text{a}}+\vec{\text{b}}$ $\overrightarrow{\text{BC}}=(\lambda+1)\vec{\text{b}}$ Using $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$, we get $\overrightarrow{\text{AB}}=\Big[\frac{(\lambda+1)}{2}\Big]\Big(\overrightarrow{\text{BC}}\Big)$ Let $\Big(\frac{\lambda+1}{2}\Big)=\mu$Since $\lambda$ is a real number. So,
$\mu$ is also a real number. So, $\overrightarrow{\text{AB}}$ is parallel to $\overrightarrow{\text{BC}}$, but $\vec{\text{B}}$ is a common vector. Hence, A, B, C are collinear.
View full question & answer→Question 385 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.
AnswerGiven points A(1, -2, -8), B(5, 0, -2), C(11, 3, 7).
Therefore, $\overrightarrow{\text{AB}}=5\hat{\text{i}}+0\hat{\text{j}}-2\hat{\text{k}}-\hat{\text{i}}+2\hat{\text{j}}+8\hat{\text{k}}=4\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$
$\overrightarrow{\text{BC}}=11\hat{\text{i}}+3\hat{\text{j}}+7\hat{\text{k}}-5\hat{\text{i}}+2\hat{\text{k}}=6\hat{\text{i}}+3\hat{\text{j}}+9\hat{\text{k}}$
and, $\overrightarrow{\text{AC}}=11\hat{\text{i}}+3\hat{\text{j}}+7\hat{\text{k}}-\hat{\text{i}}+2\hat{\text{j}}+8\hat{\text{k}}=10\hat{\text{i}}+5\hat{\text{j}}+15\hat{\text{k}}$
Clearly, $\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}$
Hence A, B, C are collinear.
Suppose B divides AC in the ratio $\lambda:1$. Then the position vector B is
$\Big(\frac{11\lambda+1}{\lambda+1}\Big)\hat{\text{i}}+\Big(\frac{3\lambda-2}{\lambda+1}\Big)\hat{\text{j}}+\Big(\frac{7\lambda-8}{\lambda+1}\Big)\hat{\text{k}}$
But the position vector of B is $5\hat{\text{i}}+0\hat{\text{j}}-2\hat{\text{k}}$.
$\Big(\frac{11\lambda+1}{\lambda+1}\Big)=5,\Big(\frac{3\lambda-2}{\lambda+1}\Big)=0,\Big(\frac{7\lambda-8}{\lambda+1}\Big)=-2$
$\Rightarrow11\lambda+1=5\lambda+5,\ 3\lambda-2=0,\ 7\lambda-8=-2\lambda-2$
$\Rightarrow6\lambda=4,\ 3\lambda=2,\ 9\lambda=6$
$\Rightarrow\lambda=\frac{2}3,\ \lambda=\frac{2}3,\ \lambda=\frac{2}3$
View full question & answer→Question 395 Marks
Show that the points $2\hat{\text{i}},-\hat{\text{i}}-4\hat{\text{j}}\text{ and }-\hat{\text{i}}+4\hat{\text{j}}$ form an isosceles triangle.
AnswerGiven:- The points A, B, C with position vectors $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ respectively.
Also, $\vec{\text{a}}=2\hat{\text{i}}$
$\vec{\text{b}}=-\hat{\text{i}}-4\hat{\text{j}}$
$\vec{\text{c}}=-\hat{\text{i}}+4\hat{\text{j}}$
Then,
$\overrightarrow{\text{AB}}=\vec{\text{b}}-\vec{\text{a}}$
$\Rightarrow\overrightarrow{\text{AB}}=\big(-\hat{\text{i}}-4\hat{\text{j}}\big)-2\hat{\text{i}}$
$\Rightarrow\overrightarrow{\text{AB}}=-3\hat{\text{i}}-4\hat{\text{j}}$
Now, $\Big|\overrightarrow{\text{AB}}\Big|=\sqrt{(-3)^2+(-4)^2}$
$=\sqrt{9+16}$
$=\sqrt{25}$
$=5$
$\overrightarrow{\text{BC}}=\vec{\text{c}}-\vec{\text{b}}$
$\Rightarrow\overrightarrow{\text{BC}}=\big(-\hat{\text{i}}+4\hat{\text{j}}\big)-\big(-\hat{\text{i}}-4\hat{\text{j}}\big)$
$\Rightarrow\overrightarrow{\text{BC}}=-\hat{\text{i}}+4\hat{\text{j}}+\hat{\text{i}}+4\hat{\text{j}}$
$\Rightarrow\overrightarrow{\text{BC}}=8\hat{\text{j}}$
View full question & answer→Question 405 Marks
Using vector method, prove that the point is collinear:
A(6, -7, -1), B(2, -3, 1) and C(4, -5, 0)
AnswerGiven the points A(6, -7, -1), B(2, -3, 1) and C(4, -5, 0). Then, $\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A$=2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}-6\hat{\text{i}}+7\hat{\text{j}}+\hat{\text{k}}$
$=-4\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}}$
$=-2\big(2\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}}\big)$$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=4\hat{\text{i}}-5\hat{\text{j}}-2\hat{\text{k}}+3\hat{\text{j}}-\hat{\text{k}}$
$=2\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}}$
$\therefore\ \overrightarrow{\text{AB}}=-2\overrightarrow{\text{BC}}$
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are parallel vectors. But B is a point common to them. Hence, the given points A, B, and C are collinear.
View full question & answer→Question 415 Marks
If $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are non-coplanar vectors, prove that the point having the following position vectors is collinear:$\vec{\text{a}},\ \vec{\text{b}},\ 3\vec{\text{a}}-2\vec{\text{b}}$
AnswerLet the points be A, B, C
Position vector of $\text{A}=\vec{\text{a}}$
Position vector of $\text{B}=\vec{\text{b}}$
Position vector of $\text{C}=3\vec{\text{a}}-2\vec{\text{b}}$
$\overrightarrow{\text{AB}}=$Position vector of B - Postion vector of A
$=\vec{\text{b}}+\vec{\text{a}}$
$\overrightarrow{\text{BC}}=$Position vector of C - Postion vector of B
$=3\vec{\text{a}}-2\vec{\text{b}}-\vec{\text{b}}$
$=3\vec{\text{a}}-3\vec{\text{b}}$
Using $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$
Let $\overrightarrow{\text{BC}}=\lambda\Big(\overrightarrow{\text{AB}}\Big)$ [where $\lambda$ is and scalar]
$3\vec{\text{a}}-3\vec{\text{b}}=\lambda\big(\vec{\text{b}}-\vec{\text{a}}\big)$
$3\vec{\text{a}}-3\vec{\text{b}}=\lambda\vec{\text{b}}-\lambda\vec{\text{a}}$
$3\vec{\text{a}}-3\vec{\text{b}}=\lambda\vec{\text{a}}+\lambda\vec{\text{b}}$
Comparing the coefficients of LHS and RHS, we get,
$-\lambda=3$
$\lambda=3$
$\lambda=-3$
Since the value of $\lambda$ are different.
Therefore,
A, B, C are not collinear.
View full question & answer→Question 425 Marks
If P is a point and ABCD is a quadrilateral and $\overrightarrow{\text{AP}}+\overrightarrow{\text{PB}}+\overrightarrow{\text{PD}}=\overrightarrow{\text{PC}}$, show that ABCD is a parallelogram.
AnswerGiven: ABCD is a quadrilateral such that $\overrightarrow{\text{AP}}+\overrightarrow{\text{PB}}+\overrightarrow{\text{PD}}=\overrightarrow{\text{PC}}$.To show: ABCD is a parallelogram.
Proof: Consider,
$\overrightarrow{\text{AP}}+\overrightarrow{\text{PB}}+\overrightarrow{\text{PD}}=\overrightarrow{\text{PC}}$
$\Rightarrow\overrightarrow{\text{AP}}+\overrightarrow{\text{PB}}=\overrightarrow{\text{PC}}-\overrightarrow{\text{PD}}$
$\Rightarrow\overrightarrow{\text{AB}}=\overrightarrow{\text{DC}}$ $\Big[\because\ \overrightarrow{\text{AP}}+\overrightarrow{\text{PB}}=\overrightarrow{\text{AB}}\text{ and }\overrightarrow{\text{PD}}+\overrightarrow{\text{DC}}=\overrightarrow{\text{PC}}\Big]$
Again,
$\overrightarrow{\text{AP}}+\overrightarrow{\text{PB}}+\overrightarrow{\text{PD}}=\overrightarrow{\text{PC}}$
$\Rightarrow\overrightarrow{\text{AP}}+\overrightarrow{\text{PD}}=\overrightarrow{\text{PC}}-\overrightarrow{\text{PB}}$
$\Rightarrow\overrightarrow{\text{AD}}=\overrightarrow{\text{BC}}$ $\Big[\because\ \overrightarrow{\text{AP}}+\overrightarrow{\text{PD}}=\overrightarrow{\text{AD}}\text{ and }\overrightarrow{\text{PB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{PC}}\Big]$
Since, opposite sides of the quadrilateral are equal and parallel.
Hence, ABCD is a parallelogram.
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