Question 1012 Marks
For any two non-zero vectors, write the value of $\frac{\big|\vec{\text{a}}+\vec{\text{b}}\big|^2+\big|\vec{\text{a}}-\vec{\text{b}}\big|^2}{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2}.$
AnswerWe have
$\frac{\big|\vec{\text{a}}+\vec{\text{b}}\big|^2+\big|\vec{\text{a}}-\vec{\text{b}}\big|^2}{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2}$
$=\frac{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}+|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2}$
$=\frac{2\big(|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2\big)}{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2}$
$=2$
View full question & answer→Question 1022 Marks
Define vector product of two vectors.
AnswerIf $\vec{\text{a}}$ and $\vec{\text{b}}$ are two non-zero non-parallel vectors, then the vector product denoted by $\vec{\text{a}}\times\vec{\text{b}}$ is defined as $\vec{\text{a}}\times\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\eta.$
Here, $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ and $\eta$ is
the unit vector perpendicular to the plane of $\vec{\text{a}}$ and $\vec{\text{b}}$ such that $\vec{\text{a}},\vec{\text{b}}$ and $\eta$ from a right handed system.
View full question & answer→Question 1032 Marks
Find the magnitude of two vectors $\vec{a}\ \text{and}\ \vec{b},$ having the same magnitude and such that the angle between them is 60° and their scalar product is $\frac{1}{2}\cdot$
Answer$\text{Given:}\ \ \big|\vec{a}\big|=\Big|\vec{b}\Big|, \text{angle}\ \theta$ (say) between $\vec{a}\ \text{and}\ \vec{b},$ is 60° and their scalar (i.e., dot) product $=\frac{1}{2}$ $\Rightarrow\ \ \vec{a}.\vec{b}=\frac{1}{2}\ \ \Rightarrow\ \ \big|\vec{a}.\Big|\vec{b}\Big|\ \text{cos}\ \theta=\frac{1}{2}$ $\text{Putting}\ \big|\vec{a}\big|=\Big|\vec{b}\Big| \ \text{and}\ \theta=60^\circ,\ \text{we have}\ \ \ \big|\vec{a}\big|.\big|\vec{a}\big|\ \text{cos}\ 60^\circ=\frac{1}{2}$ $\Rightarrow\ \ \big|\vec{a}|^2.\Big(\frac{1}{2}\Big)=\frac{1}{2}\ \ \Rightarrow\ \ \big|\vec{a}\big|^2=1\ \ \Rightarrow\ \ \big|\vec{a}\big|=1$$\therefore\ \ \Big|\vec{b}\Big|=\big|\vec{a}\big|=1$
$\therefore\ \ \big|\vec{a}\big|=1\ \text{and}\ \Big|\vec{b}\Big|=1$
View full question & answer→Question 1042 Marks
Find the position vector of the min-point of the line segment AB, where A is the point (3, 4, -2) and B is the point (1, 2, 4).
AnswerGiven: A(3, 4, -2) and B(1, 2, 4) Let C is the mid-point of AB $\therefore$ Position vector of $\text{C}=\frac{3\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}+\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}}2$$=\frac{4\hat{\text{i}}+6\hat{\text{j}}+2\hat{\text{k}}}2$
$=2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$
View full question & answer→Question 1052 Marks
If $\vec{\text{a}}=3\hat{\text{i}}+4\hat{\text{j}}$ and $\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},$ find the value of $\big|\vec{\text{a}}\times\vec{\text{b}}\big|.$
AnswerGiven:
$\vec{\text{a}}=3\hat{\text{i}}+4\hat{\text{j}}+0\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
$\therefore\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&4&0\\1&1&1 \end{vmatrix}$
$=(4-0)\hat{\text{i}}-(3-0)\hat{\text{j}}+(3-4)\hat{\text{k}}$
$=4\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{16+9+1}$
$=\sqrt{26}$
View full question & answer→Question 1062 Marks
If $\vec{\text{c}}$ is a unit vector perpendicular to the vectors $\vec{\text{a}}$ and $\vec{\text{b}},$ write another unit vector perpendicular to $\vec{\text{a}}$ and $\vec{\text{b}}.$
Answer$\vec{\text{c}}$ is a unit vector perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}.$
$\Rightarrow\vec{\text{c}}=\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\times\vec{\text{b}}\big|}$
$\Rightarrow-\vec{\text{c}}=\frac{\vec{\text{b}}\times\vec{\text{a}}}{\big|\vec{\text{a}}\times\vec{\text{b}}\big|}$
Therefore, $-\vec{\text{c}}$ is perpendicular to $\vec{\text{b}}$ and $\vec{\text{a}}.$
Thus, $-\vec{\text{c}}$ is another unit vector perpendicular to $\vec{\text{a}}$ and $\vec{\text{b}}.$
View full question & answer→Question 1072 Marks
Find the angle between the vectors $\vec{\text{a}} $ and $\vec{\text{b}},$ where$\vec{\text{a}}=\hat {\text{i}}-\hat{\text{j}}$ and $\vec{\text{b}} = \hat{\text{j}}+\hat{\text{k}}$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.
$\big|\vec{\text{a}}\big|=\sqrt{(1)^2+(-1)^2}=\sqrt{2}$
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2+(1)^2}=\sqrt{2}$
$\vec{\text{a}}.\vec{\text{b}}=0-1+0=-1$
$\cos\theta=\frac{\vec{\text{a }}.\vec{\text{ b}}}{\big|\vec{a}\big|\big|\vec{b}\big|}=\frac{-1}{\sqrt{2}\sqrt{2}}=\frac{-1}{2}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{-1}{2}\big)=\frac{2\pi}{3}$
View full question & answer→Question 1082 Marks
Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a}=\hat{i}-\hat{j}+3\hat{k}\ \text{and}\ \vec{b}=2\hat{i}-7\hat{j}+\hat{k}.$
Answer Given: Vectors representing two adjacent sides of a parallelogram are $\vec{a}=\hat{i}-\hat{j}+3\hat{k}\ \text{and}\ \vec{b}=2\hat{j}-7\hat{j}+\hat{k}$ $\therefore\ \vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-1&3\\2&-7&1\end{vmatrix}=\hat{i}(-1+21)-\hat{j}(1-6)+\hat{k}(-7+2)$ $=20\hat{i}+5\hat{j}-5\hat{k}$ $\text{Now}\ \ \text{Area of parallelogram}=\big|\vec{a}\times\vec{b}\big|=\sqrt{400+25+25}$ $=\sqrt{450}=15\sqrt{2}\ \text{sq. units}$
View full question & answer→Question 1092 Marks
Show that the following triads of vectors are coplanar:
$\vec{\text{a}}=-4\hat{\text{i}}-6\hat{\text{j}}-2\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+4\hat{\text{j}}+3\hat{\text{k}},\vec{\text{c}}=-8\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$
AnswerWe know that three vectors $\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}$ are coplanar iff their scalar triple product is zero i.e. $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0.$
Here,
$\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=\begin{vmatrix}-4&-6&-2\\-1&4&3\\-8&-1&3 \end{vmatrix}\\$
$=-4(12+3)+6(-3+24)-2(1+32)$
$= -60 + 126 -66$
$= 0$
Hence, the given vectors are coplanar.
View full question & answer→Question 1102 Marks
Write the angle between the vectors $\vec{\text{a}}\times\vec{\text{b}}$ and $\vec{\text{b}}\times\vec{\text{a}}.$
Answer$\vec{\text{b}}\times\vec{\text{a}}=-\vec{\text{a}}\times\vec{\text{b}}$
So, $\vec{\text{a}}\times\vec{\text{b}}$ and $\vec{\text{b}}\times\vec{\text{a}}$ are vectors of same magnitude but opposite in direction.
Thus, the angle between the vectors $\vec{\text{a}}\times\vec{\text{b}}$ and $\vec{\text{b}}\times\vec{\text{a}}$ is $180^\circ.$
View full question & answer→Question 1112 Marks
Write a unit vector making equal acute angles with the coordinates axes.
AnswerSuppose $\vec{\text{r}}$ makes an angle $\alpha$ wuth each of the axis OX, OY and OZ.
Then, its direction cosines are $\text{l}=\cos\alpha,\ \text{m}=\cos\alpha,\ \text{n}=\cos\alpha$.
Now,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\ \text{l}^2+\text{l}^2+\text{l}^2=1$ $[\because\text{l = m = n}]$
$\Rightarrow\ 3\text{l}^2=1$
$\Rightarrow\ \text{l}^2=\frac{1}3$
$\Rightarrow\ \text{l}=\pm\frac{1}{\sqrt3}$
Since the angle is acute Hence, we take only positive value
Therefore, unit vector is $\Big(\frac{1}{\sqrt3}\hat{\text{i}}+\frac{1}{\sqrt3}\hat{\text{j}}+\frac{1}{\sqrt3}\hat{\text{k}}\Big)$.
View full question & answer→Question 1122 Marks
Find the vector in the direction of vector $2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}$ which has magnitude 21 units.
AnswerLet $\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}$
$\therefore\ |\vec{\text{a}}|=\sqrt{2^2+(-3)^2+6^2}$
$=\sqrt{4+9+36}$
$=\sqrt{49}$
$=7$
Unit vector in the direction of $\vec{\text{a}}=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}}7$
$\therefore$ vector in the direction of vector $\vec{\text{a}}$ which has magnitude 21 units
$=21\times\Big(\frac{2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}}7\Big)$
$=3\big(2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}\big)$
$=6\hat{\text{i}}-9\hat{\text{j}}+18\hat{\text{k}}$
View full question & answer→Question 1132 Marks
Write the length (magnitude) of a vector whose projections on the coordinate axes are 12, 3 and 4 units.
Answer Given: Projections on the coordinate axes are 12, 3, 4 units. Therefore,
Length of vector $=\sqrt{12^2+3^2+4^2}$
$=\sqrt{169}$
$=13$
View full question & answer→Question 1142 Marks
Write the value of $\big(\hat{\text{i}}\times\hat{\text{j}}\big).\hat{\text{k}}+\big(\hat{\text{j}}+\hat{\text{k}}\big).\hat{\text{j}}$
Answer$\big(\hat{\text{i}}\times\hat{\text{j}}\big).\hat{\text{k}}+\big(\hat{\text{j}}+\hat{\text{k}}\big).\hat{\text{j}}$
$=\hat{\text{k}}.\hat{\text{k}}+\hat{\text{j}}.\hat{\text{j}}+\hat{\text{k}}.\hat{\text{j}}$ $\big(\because\hat{\text{i}}\times\hat{\text{j}}=\hat{\text{k}}\big)$
$=|\hat{\text{k}}|^2+|\hat{\text{j}}|^2+0$ $\big(\because\hat{\text{k}}.\hat{\text{j}}=0\big)$
$=1^2+1^2$
$=2$
View full question & answer→Question 1152 Marks
If $|\vec{\text{a}}|=4$ and $-3\leq\lambda\leq2$, then write the range of $|\lambda\vec{\text{a}}|$.
AnswerIt is given that
$-3\leq\lambda\leq2$
$\Rightarrow\ -3\times|\vec{\text{a}}|\leq\lambda|\vec{\text{a}}|\leq2\times|\vec{\text{a}}|$
$\Rightarrow\ -3\times4\leq|\lambda\vec{\text{a}}|\leq2\times4$ $\big(\text{k}|\vec{\text{a}}|=|\text{k}\vec{\text{a}}|,\ \text{k}$ is scalar$\big)$
$\Rightarrow\ -12\leq|\lambda\vec{\text{a}}|\leq8$
Thus, the range of $|\lambda\vec{\text{a}}|$ is [-12, 8]
View full question & answer→Question 1162 Marks
If $|\vec{\text{a}}|=10,\big|\vec{\text{b}}\big|=2$ and $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=16,$ find $\vec{\text{a}}.\vec{\text{b}}.$
AnswerWe know
$\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$
$\Rightarrow\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+(16)^2=(10)^2\times2^2$ $\big(\because\big|\vec{\text{a}}\times\vec{\text{b}}\big|=16,|\vec{\text{a}}|=10\text{ and }\big|\vec{\text{b}}\big|=2\big)$
$\Rightarrow\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+256=400$
$\Rightarrow\big(\vec{\text{a}}.\vec{\text{b}}\big)^2=144$
$\Rightarrow\big(\vec{\text{a}}.\vec{\text{b}}\big)^2=\pm12$
View full question & answer→Question 1172 Marks
Find the projection of the vector $\hat{i}+3\hat{j}+7\hat{k}$ on the vector $7\hat{i}-\hat{j}+8\hat{k}.$
Answer$\text{Let}\ \ \vec{a}=\hat{i}+3\hat{j}+7\hat{k}\ \text{and}\ \vec{b}=7\hat{i}-\hat{j}+8\hat{k}$
Porjection of vector $\vec{a}\ \text{and}\ \vec{b}=\frac{\vec{a}.\vec{b}}{\big|\vec{b}\big|}=\frac{(1)(7)+(3)(-1)+7(8)}{\sqrt{(7)^2+(-1)^2+(8)^2}}$ $=\frac{7-3+56}{\sqrt{49+1+64}}=\frac{60}{\sqrt{114}}$
View full question & answer→Question 1182 Marks
Find the position vector of the mid point of the vector joining the points P(2, 3, 4) and Q(4, 1, –2).
AnswerThe position vector of mid-point R of the vector joining points P(2, 3, 4) and Q(4, 1, -2) is given by,
$\overrightarrow{\text{OR}}=\frac{\big(2\hat{i}+3\hat{j}+4\hat{k}\big)+\big(4\hat{i}+\hat{j}-2\hat{k}\big)}{2}$ $=\frac{(2+4)\hat{i}+(3+1)\hat{j}+(4-2)\hat{k}}{2}$
$=\frac{6\hat{i}+4\hat{j}+2\hat{k}}{2}=3\hat{i}+2\hat{j}+\hat{k}$
View full question & answer→Question 1192 Marks
Find the values of 'a' for which the vectors
$\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}},\vec\beta={\text{a}}\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$ and $\vec\gamma=\hat{\text{i}}+2\hat{\text{j}}+\text{a}\hat{\text{k}}$ are coplanar.
AnswerGiven:
$\vec\alpha=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
$\vec\beta={\text{a}}\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
$\vec\gamma=\hat{\text{i}}+2\hat{\text{j}}+\text{a}\hat{\text{k}}$
We know that three vectors $\vec\alpha,\vec\beta,\vec\gamma$ are coplanar if their scalar product is zero.
$\therefore\big[\vec\alpha \vec \beta \vec\gamma\big]=0$
$\begin{vmatrix}1&2&1\\\text{a}&1&2\\1&2&\text{a} \end{vmatrix}=0$
$\Rightarrow1(\text{a}-4)-2(\text{a}^2-2)+1(2\text{a}-1)=0$
$\Rightarrow-2\text{a}^2+3\text{a}-1=0$
$\Rightarrow2\text{a}^2-3\text{a}+1=0$
$\Rightarrow(\text{a}-1)(2\text{a}-1)=0$
$\Rightarrow\text{a}=1,\frac{1}{2}$
View full question & answer→Question 1202 Marks
For what vaiue of $\lambda$ are the vectors $\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$ perpendicular to each other?
AnswerWe know
$\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
Given, $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular. so, their dot product is zero.
$\Rightarrow\big(2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}\big).\big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big)=0$
$\Rightarrow2-2\lambda+3=0$
$\Rightarrow-2\lambda+5=0$
$\therefore\lambda=\frac{5}{2}$
View full question & answer→Question 1212 Marks
Find the unit vector in the direction of vector $\overrightarrow{\text{PQ}}$ where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.
AnswerThe given points are P (1, 2, 3) and Q (4, 5, 6).$\therefore\overrightarrow{\text{PQ}}=(4-1)\hat{i}+(5-2)\hat{j}+(6-3)\hat{k}=3\hat{i}+3\hat{j}+3\hat{k}$
$\Bigg|\overrightarrow{\text{PQ}}\Bigg|=\sqrt{3^2+3^2+3^2}=\sqrt{9+9+9}=\sqrt{27}=3\sqrt{3} $
Hence, the unit vector in the direction of $\overrightarrow{\text{PQ}}$ is $\frac{\overrightarrow{\text{PQ}}}{\Big|\overrightarrow{\text{PQ}}\Big|}=\frac{3\hat{i}+3\hat{j}+3\hat{k}}{3\sqrt{3}}=\frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\frac{1}{\sqrt{3}}\hat{k}$
View full question & answer→Question 1222 Marks
Write the unit vector in the direction of $\vec{\text{a}}=3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$.
AnswerWe have,$\vec{\text{a}}=3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$
$|\vec{\text{a}}|=\sqrt{3^2+(-2)^2+6^2}$ $=\sqrt{9+4+36}$ $=\sqrt{49}$ $=7$ $\therefore$ Unit vector in the direction of $\vec{\text{a}}=\hat{\text{a}}=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{1}7\big(3\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}\big)=\frac{3}7\hat{\text{i}}-\frac{2}7\hat{\text{j}}+\frac{6}7\hat{\text{k}}$
View full question & answer→Question 1232 Marks
Prove that $\big(\vec{\text{a}}+\vec{\text{b}}\big)\cdot\big(\vec{\text{a}}+\vec{\text{b}}\big)=\big|\vec{\text{a}}\big|^2+\Big|\vec{\text{b}}\Big|^2,$ if and only if $\vec{\text{a}},\vec{\text{b}}$ are perpendicular, given $\vec{\text{a}}\neq\vec{\text{0}},\vec{\text{b}}\neq\vec{\text{0}}.$
Answer$\Big(\vec{\text{a}}+\vec{\text{b}}\Big)\cdot\Big(\vec{\text{a}}+\vec{\text{b}}\Big)=\big|\vec{\text{a}}\big|^2+\Big|\vec{\text{b}}\Big|^2$
$\Leftrightarrow\vec{\text{a}}\cdot\vec{\text{a}}+\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{a}}+\vec{\text{b}}\cdot\vec{\text{b}}=\big|\vec{\text{a}}\big|^2+\Big|\vec{\text{b}}\Big|^2$ [Distributivity of scalar products over addition]
$\Leftrightarrow\big|\vec{\text{a}}\big|^2+2\vec{\text{a}}\cdot\vec{\text{b}}+\Big|\vec{\text{b}}\Big|^2=\big|\vec{\text{a}}\big|^2+\Big|\vec{\text{b}}\Big|^2\ \ \ $ $\Big[\vec{\text{a}}\cdot\vec{\text{b}}=\vec{\text{b}}\cdot\vec{\text{a}}\ \text{(Scalar product is commutative)}\Big]$
$\Leftrightarrow2\vec{\text{a}}\cdot\vec{\text{b}}=0$
$\Leftrightarrow\vec{\text{a}}\cdot\vec{\text{b}}=0$
$\therefore\vec{\text{a}}\ \text{and}\ \vec{\text{b}} \ \text{are perpendicular.}$ $\ \ \ \Big[\vec{\text{a}}\neq\vec{\text{0}},\ \vec{\text{b}}\neq\vec{\text{0}}\ \text{(Given)}\Big]$
View full question & answer→Question 1242 Marks
Find $\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big), $ if $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$.
AnswerThe given vectors are $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{j}}+3\vec{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
Now,
$\vec{\text{b}}\times\vec{\text{c}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\-1&2&1\\3&1&2 \end{vmatrix}=3\hat{\text{i}}+5\hat{\text{j}}-7\hat{\text{k}}$
$\therefore\vec{\text{a}}\big(\vec{\text{b}}\times\vec{\text{c}}\big)=\big(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big).\big(3\hat{\text{i}}+5\hat{\text{j}}-7\hat{\text{k}}\big)$
$=2\times3+1\times5=3\times(-7)$
$=6+5-21=-10$
View full question & answer→Question 1252 Marks
Find the values of x and y so that the vectors $2\hat{i}+3\hat{j}\ \text{and}\ x\hat{i}\ +y\hat{j}$ are equal.
AnswerThe two vectors $2\hat{i}+3\hat{j}\ \text{and}\ x\hat{i}\ +y\hat{j}$ will be equal if their corresponding components are equal.
Hence, the required values of x and y are 2 and 3 respectively.
View full question & answer→Question 1262 Marks
Find the value of x for which $\text{x}\Big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)$ is a unit vector.
Answer$\text{x}\Big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)\ \text{is a unit vector if}\ \Big|\text{x}\Big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)\Big|=1$Now,
$\Big|\text{x}\Big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)\Big|=1$ $\Rightarrow{}\sqrt{\text{x}^{2}+\text{x}^{2}+\text{x}^{2}}=1$ $\Rightarrow\sqrt{3\text{x}^2}=1$ $\Rightarrow\sqrt{3\text{x}}=1$ $\Rightarrow\text{x}=\pm\frac{1}{\sqrt{3}}$Hence, the required value of x is $\pm\frac{1}{\sqrt{3}}$
View full question & answer→Question 1272 Marks
If $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are mutually perpendicular unit vectors, write the value of $\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|.$
AnswerGiven that $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are unit vectors.
So, $|\vec{\text{a}}|=1,\big|\vec{\text{b}}\big|=1$ and $|\vec{\text{c}}|=1\dots(1)$
Since they are mutually perpendicular,
$\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0\dots(2)$
Now,
$\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|+|\vec{\text{c}}|^2+2\vec{\text{a}}.\vec{\text{b}}+2\vec{\text{b}}.\vec{\text{c}}+2\vec{\text{c}}.\vec{\text{a}}$
$=1+1+1+0+0+0$ [using (1) and (2)]
$=3$
$\therefore\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=\sqrt{3}$
View full question & answer→Question 1282 Marks
If $\vec{\text{a}}$ is a unit vector such that $\vec{\text{a}}\times\hat{\text{i}}=\hat{\text{j}},$ find $\vec{\text{a}}.\hat{\text{i}}.$
AnswerWe know
$\hat{\text{k}}\times\hat{\text{i}}=\hat{\text{j}}\dots(1)$
Given: $\vec{\text{a}}\times\hat{\text{i}}=\hat{\text{j}}\dots(2)$
Comparing (1) and (2), we get
$\vec{\text{a}}=\hat{\text{k}}$
Now,
$\vec{\text{a}}.\hat{\text{i}}=\hat{\text{k}}.\hat{\text{i}}$
$=0$
View full question & answer→Question 1292 Marks
$\text{Find}\ |\vec{a}\ \times\vec{b}|,\ \text{if}\ \vec{a}=\hat{i}-7\hat{j}+7\hat{k}\ \text{and}\ \vec{b}=3\hat{i}-2\hat{j}+2\hat{k}.$
Answer$\text{Give:}\ \ \vec{a}=\hat{i}-7\hat{j}+7\hat{k}\ \text{and}\ \vec{b}=3\hat{i}-2\hat{j}+2\hat{k}$
$\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-7&7\\3&-2&2\end{vmatrix}$
Expanding along first row,
$\vec{a}\times\vec{b}=\hat{i}\begin{vmatrix}-7&7\\-2&2\end{vmatrix}-\hat{j}\begin{vmatrix}1&7\\3&2\end{vmatrix}+\hat{k}\begin{vmatrix}1&-7\\3&-2\end{vmatrix}$ $=\hat{i}(-14+14)-\hat{j}(2-21)+\hat{k}(-2+21)$
$=0\hat{i}+19\hat{j}+19\hat{k}$
$\therefore\ \ \Big|\vec{a}\times\vec{b}\Big|=\sqrt{(0)^2+(19)^2+(19)^2}=\sqrt{2(19)^2}=19\sqrt{2}$
View full question & answer→Question 1302 Marks
Write the value $\big(\hat{\text{i}}\times\hat{\text{j}}\big).\hat{\text{k}}+\hat{\text{i}}.\hat{\text{j}}$
Answer $\big(\hat{\text{i}}\times\hat{\text{j}}\big).\hat{\text{k}}+\hat{\text{i}}.\hat{\text{j}}$$=\hat{\text{k}}.\hat{\text{k}}+0$
$=|\hat{\text{k}}|^2+0$
$=1^2+0$ $\big(\because|\text{k}|=1\big)$
$=1$
View full question & answer→Question 1312 Marks
If P, Q and R are three collinear points such that $\overrightarrow{\text{PQ}}=\vec{\text{a}}\text{ and }\overrightarrow{\text{QR}}=\vec{\text{b}}$. Find the vector $\overrightarrow{\text{PR}}$.
Answer Given that, P, Q, R are collinear.It also given that, $\overrightarrow{\text{PQ}}=\vec{\text{a}}\text{ and }\overrightarrow{\text{QR}}=\vec{\text{b}}$
$\overrightarrow{\text{PR}}=\overrightarrow{\text{PQ}}+\overrightarrow{\text{QR}}$
$=\vec{\text{a}}+\vec{\text{b}}$
$\overrightarrow{\text{PR}}=\vec{\text{a}}+\vec{\text{b}}$
View full question & answer→Question 1322 Marks
Write the value of $\hat{\text{i}}\times\big(\hat{\text{j}}+\hat{\text{k}}\big)+\hat{\text{j}}\times\big(\hat{\text{k}}+\hat{\text{i}}\big)+\hat{\text{k}}\times\big(\hat{\text{i}}+\hat{\text{j}}\big).$
Answer$\hat{\text{i}}\times\big(\hat{\text{j}}+\hat{\text{k}}\big)+\hat{\text{j}}\times\big(\hat{\text{k}}+\hat{\text{i}}\big)+\hat{\text{k}}\times\big(\hat{\text{i}}+\hat{\text{j}}\big)$
$=\big(\hat{\text{i}}\times\hat{\text{j}}\big)+\big(\hat{\text{i}}\times\hat{\text{k}}\big)+\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\big(\hat{\text{j}}\times\hat{\text{i}}\big)+\big(\hat{\text{k}}\times\hat{\text{i}}\big)+\big(\hat{\text{k}}\times\hat{\text{j}}\big)$
$=\hat{\text{k}}-\hat{\text{j}}+\hat{\text{i}}-\hat{\text{k}}+\hat{\text{j}}-\hat{\text{i}}=\vec{0}$
View full question & answer→Question 1332 Marks
If $|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=5$ and $\vec{\text{a}}.\vec{\text{b}}=2,$ find $\big|\hat{\text{a}}-\hat{\text{b}}\big|.$
AnswerWe have $|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=5$ and $\vec{\text{a}}.\vec{\text{b}}=2$
Now,$\big|\hat{\text{a}}-\hat{\text{b}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}.\vec{\text{b}}$
$=2^2+5^2-2(2)$
$=4+25-4$
$=25$
$\therefore\big|\vec{\text{a}}-\vec{\text{b}}\big|=\sqrt{25}=5$
View full question & answer→Question 1342 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors such that $\big|\vec{\text{a}}.\vec{\text{b}}\big|=\big|\vec{\text{a}}\times\vec{\text{b}}\big|,$ write the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
We know
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\big|\sin\theta\big|$
$\big|\vec{\text{a}}.\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\big|\cos\theta\big|$
Now,
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\big|\vec{\text{a}}.\vec{\text{b}}\big|$ (Given)
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big||\sin\theta|=|\vec{\text{a}}|\big|\vec{\text{b}}\big||\cos\theta|$
$\Rightarrow|\sin\theta|=|\cos\theta|$
$\Rightarrow\theta=\frac{\pi}{4}$
View full question & answer→Question 1352 Marks
Let the vectors $\vec{a},\vec{b},\vec{c}$ be given as $a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k},\ b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k},$ $c_{1}\hat{i}+c_{2}\hat{j}+c_{3}\hat{k}.$ Then show that $\vec{a}\times(\vec{b}+\vec{c})=\vec{a}\times\vec{b}+\vec{a}\times\vec{c}.$
AnswerGiven: $\text{Vector}\ \vec{a}=a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k},\ \vec{b}=b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k}$ $\text{and}\ \vec{c}=c_{1}\hat{i}+c_{2}\hat{j}+c_{3}\hat{k}$
$\therefore\ \ \vec{b}+\vec{c}=(b_1+c_1)\hat{i}+(b_2+c_2)\hat{j}+(b_3+c_3)\hat{k}$
$\text{Now}\ \ \ \ \text{L.H.S}=\vec{a}\times\big(\vec{b}+\vec{c}\big)= \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ a_1& a_2 & a_3 \\ b_1+c_1& b_2+c_2 & b_3+c_3 \end{vmatrix}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ a_1& a_2 & a_3 \\ b_1& b_2& b_3\end{vmatrix}+\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ a_1& a_2 & a_3 \\ c_1& c_2& c_3\end{vmatrix}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{[By Property of Determinants]}$
$=\vec{a}\times\vec{b}+\vec{a}\times\vec{c}=\text{R.H.S}.$
View full question & answer→Question 1362 Marks
Find the scalar components and magnitude of the vector joining the points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$.
AnswerThe vector joining the points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ can be obtained by,
$\overrightarrow{\text{PQ}}$ = Position vector of Q - Position vector of P
$=(\text{x}_{2}-\text{x}_{1})\hat{\text{i}}+(\text{y}_{2}-\text{y}_{1})\hat{\text{j}}+(\text{z}_{2}-\text{z}_{1})\hat{\text{k}}$
$\bigg|\overrightarrow{\text{PQ}}\bigg|=\sqrt{(\text{x}_{2}-\text{x}_{1})^{2}+(\text{y}_{2}-\text{y}_{1})^{2}+(\text{z}_{2}-\text{z}_{1})^{2}}$
Hence, the scalar components and the magnitude of the vector joining the given points are respectively $ \{(\text{x}_{2}-\text{x}_{1}),(\text{y}_{2}-\text{y}_{1}),(\text{z}_{2}-\text{z}_{1})\}\ \text{and}$ $\sqrt{(\text{x}_{2}-\text{x}_{1})^{2}+(\text{y}_{2}-\text{y}_{1})^{2}+(\text{z}_{2}-\text{z}_{1})^{2}}$
View full question & answer→Question 1372 Marks
Write the value of $\hat{\text{i}}.\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\hat{\text{j}}.\big(\hat{\text{k}}\times\hat{\text{i}}\big)+\hat{\text{k}}.\big(\hat{\text{j}}\times\hat{\text{i}}\big).$
Answer$\hat{\text{i}}.\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\hat{\text{j}}.\big(\hat{\text{k}}\times\hat{\text{i}}\big)+\hat{\text{k}}.\big(\hat{\text{j}}\times\hat{\text{i}}\big)$$=\hat{\text{i}}.\hat{\text{i}}+\hat{\text{j}}.\hat{\text{j}}+\hat{\text{k}}\big(-\hat{\text{k}}\big)$
$=|\hat{\text{i}}|^2+|\hat{\text{j}}|^2-|\hat{\text{k}}|^2$
$=1+1-1$ $\big(\because|\hat{\text{i}}|=1,|\hat{\text{j}}|=1\text{ and } |\hat{\text{k}}|=1\big)$
$=1$
View full question & answer→Question 1382 Marks
Show that the following triads of vectors are coplanar:
$\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}},\vec{\text{b}}=3\hat{\text{i}}+2\hat{\text{j}}+7\hat{\text{k}},\vec{\text{c}}=5\hat{\text{i}}+6\hat{\text{j}}+5\hat{\text{k}}$
AnswerWe know that three vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar iff their scalar triple product is zero i.e. $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0.$
Here,
$\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=\begin{vmatrix}1&2&-1\\3&2&7\\5&6&5 \end{vmatrix}$
$=1(1-42)-2(15-35)-1(18-10)$
$=-60+126-66$
$=0$
Hence, the given vectors are coplanar.
View full question & answer→Question 1392 Marks
If $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$and $\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$, find a unit vector parallel to $2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}$.
Answer We have, $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$and $\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
$\therefore\ 2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}=2\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)\\-\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)+3\big(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)$
$=3\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$
A unit vector parallel to $2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}$ is given by $\frac{2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}}{\big|2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}\big|}=\frac{(3\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}})}{\sqrt{3^2+(-3)^2+2^2}}$
$=\frac{(3\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}})}{\sqrt{22}}$
$=\frac{3}{\sqrt{22}}\hat{\text{i}}-\frac{3}{\sqrt{22}}\hat{\text{j}}+\frac{2}{\sqrt{22}}\hat{\text{k}}$
View full question & answer→Question 1402 Marks
Write the value of $\hat{\text{i}}\times\big(\hat{\text{j}}\times\hat{\text{k}}\big).$
Answer$\hat{\text{i}}\times\big(\hat{\text{j}}\times\hat{\text{k}}\big)$$=\hat{\text{i}}\times\hat{\text{i}}$
$=0$
View full question & answer→Question 1412 Marks
Evaluate the following:$\big[2\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat{\text{i}}\hat{\text{k}}\hat{\text{j}}\big]+\big[\hat{\text{k}}\hat{\text{j}}2\hat{\text{i}}\big]$
Answer We have,
$\big[2\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat {\text{i}}\hat{\text{k}}\hat{\text{j}}\big]+\big[\hat{\text{k}}\hat{\text{j}}2\hat{\text{i}}\big]\\=(2\hat{\text{i}}\times\hat{\text{j}}).\hat{\text{k}}+(\hat{\text{i}}\times\hat{\text{k}}).2\hat{\text{i}}$
$=2\hat{\text{k}}.\hat{\text{k}}+(-\hat{\text{j}}).\hat{\text{j}}+(-\hat{\text{i}}).2\hat{\text{i}}$
$=2-1-2$
$=-1$
Therefore, $\big[2\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat{\text{i}}\hat{\text{k}}\hat{\text{j}}\big]+\big[\hat{\text{k}}\hat{\text{j}}2\hat{\text{i}}\big]=-1$
View full question & answer→Question 1422 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors such that $\vec{\text{a}}.\vec{\text{b}}=6,|\vec{\text{a}}|=3$ and $\big|\vec{\text{b}}\big|=4.$ write the projection of $\vec{\text{a}}$ on $\vec{\text{b}}.$
AnswerWe have
$\vec{\text{a}}.\vec{\text{b}}=6$ and $\big|\vec{\text{b}}\big|=4$
The projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is
$\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}$
$=\frac{6}{4}$
$=\frac{3}{2}$
View full question & answer→Question 1432 Marks
Write a vector of magnitude 12 units which makes 45º angle with x-axis, 60º angle with y-axis and an obtuse angle with z-axis.
AnswerSuppose a vector $\vec{\text{r}}$ makes an angle 45º with OX, 60º with OY and having magnitude 12 units. $\text{l}=\cos45^{\circ}=\frac{1}{\sqrt2}$ and $\text{m}=\cos60^{\circ}=\frac{1}2$Now, $\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\ \frac{1}2+\frac{1}4+\text{n}^2=1$
$\Rightarrow\ \text{n}^2=\frac{1}4$
$\Rightarrow\ =-\frac{1}2$ $[\because$ The angle with the z-axis is obtuse$]$
Therefore,
$\vec{\text{r}}=|\vec{\text{r}}|\big(\text{l}\hat{\text{i}}+\text{m}\hat{\text{j}}+\text{n}\hat{\text{k}}\big)$
$=12\Big(\frac{1}{\sqrt2}\hat{\text{i}}+\frac{1}2\hat{\text{j}}-\frac{1}2\hat{\text{k}}\Big)$
$=6\big(\sqrt2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
View full question & answer→Question 1442 Marks
If G denots the centroid of $\triangle\text{ABC}$, then write the value of $\overrightarrow{\text{GA}}+\overrightarrow{\text{GB}}+\overrightarrow{\text{GC}}$.
AnswerLet $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ be the position vectors of the vertices A, B, C respectively. Then, the position vector of the centroid G is $\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3$
Thus,
$\overrightarrow{\text{GA}}+\overrightarrow{\text{GB}}+\overrightarrow{\text{GC}}$
$=\vec{\text{a}}-\Big(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3\Big)+\vec{\text{b}}-\Big(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3\Big)+\vec{\text{c}}-\Big(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3\Big)$
$=\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)$
$=\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)-\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)$
$=\vec0$
View full question & answer→Question 1452 Marks
Find a unit vector perpendicular to each of the vector $\vec{a}+\vec{b}\ \text{and}\ \vec{a}-\vec{b},\ \text{where}$ $\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}\ \text{and}\ \vec{b}=\hat{i}+2\hat{j}-2\hat{k}.$
Answer$\text{Given:}\ \ \vec{a}=3\hat{i}+2\hat{j}+2\hat{k}\ \text{and}\ \vec{b}=\hat{i}+2\hat{j}-2\hat{k}$
$\text{On Adding}\ \vec{c}=\vec{a}+\vec{b}=3\hat{i}+2\hat{j+2\hat{k}}\ +\ \hat{i}+2\hat{j}-2\hat{k}$ $=4\hat{i}+4\hat{j}+0\hat{k}$
$\text{On Subtracting}\ \ \ \vec{d}=\vec{a}-\vec{b}=3\hat{i}+2\hat{j}+2\hat{k}\ - \ \hat{i}-2\hat{j}+2\hat{k}$ $=2\hat{i}+0\hat{j}+4\hat{k}$
$\text{Therefore,}\ \ \ \vec{n}=\vec{c}\times\vec{d}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\4&4&0\\2&0&4 \end{vmatrix}$
Expanding along first row $=\hat{i}(16-0)-\hat{i}(16-0)+\hat{k}(0-8)$
$\Rightarrow\ \ \vec{n}=16\hat{i}-16\hat{j}-8\hat{k}$
$\therefore\ \big|\vec{n}\big|=\sqrt{(16)^2+(-16)^2+(-8)^2}$ $\sqrt{256+256+64}=\sqrt{576}=24$
Therefore, a unit vector perpendicular to both $\vec{a}\ \text{and}\ \vec{b}\ \text{is}$
$\hat{n}=\pm\frac{\vec{n}}{|\vec{n}|}=\pm\frac{(16\hat{i}-16\hat{j}-8\hat{k})}{24}$
$=\pm\Big(\frac{16}{24}\hat{i}-\frac{16}{24}\hat{j}-\frac{8}{24}\hat{k}\Big)=\pm\Big(\frac{2}{3}\hat{i}-\frac{2}{3} \hat{j}-\frac{1}{3}\hat{k}\Big)$
View full question & answer→Question 1462 Marks
Write the angle between two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ with magnitudes $\sqrt{3}$ and 2 respectively if $\vec{\text{a}}.\vec{\text{b}}=\sqrt{6}.$
AnswerLet $\theta$ be the angle between$\vec{\text{a}}$ and$\vec{\text{b}}.$
Given,
$|\vec{\text{a}}|=\sqrt{3};\big|\vec{\text{b}}\big|=2;\vec{\text{a}}.\vec{\text{b}}=\sqrt{6}$
We know that
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow\sqrt{6}=(\sqrt{3})(2)\cos\theta$
$\Rightarrow\cos\theta=\frac{\sqrt{6}}{2\sqrt{3}}=\frac{1}{\sqrt{2}}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{1}{\sqrt{2}}\Big)=\frac{\pi}{4}$
View full question & answer→Question 1472 Marks
Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (–5, 7).
AnswerThe vector with the initial point P (2, 1) and terminal point Q (-5, 7) can be given by, $\overrightarrow{\text{PQ}}=(-5-2)\hat{i}+(7-1)\hat{j}$ $\Rightarrow\ \overrightarrow{\text{PQ}}=-7\hat{i}+6\hat{j}$ Hence, the required scalar components are -7 and 6 while the vector component are,$-7\hat{i}\ \text{and}\ 6\hat{j}$
View full question & answer→Question 1482 Marks
$\text{Given that}\ \vec{a}\cdot\vec{b}=0\ \text{and}\ \vec{a}\times\vec{b}=\vec{0}.$ What can you conclude about the vectors $\vec{a}\ \text{and}\ \vec{b}$?
Answer$\text{Given:}\ \ \vec{a}.\vec{b}=0\ \Rightarrow\ \ \big|\vec{a}\big|.\big|\vec{b}\big|\cos\theta=0$ $\therefore\ \ \big|\vec{a}\big|=0\ \ \text{or} \ \big|\vec{b}\big|=0$$\ \text{or}\ \cos\theta=0\ \ \Rightarrow\ \ \theta=90^\circ$ $\Rightarrow \ \ \vec{a}=0\ \ \text{or}\ \ \vec{b}=0\ \ \text{or}$ $\ \ \text{vector}\ \vec{a}\ \text{is perpendicular to}\ \vec{b}.\ \ \ \ ......\text{(i)}$ $\text{Again, given}\ \vec{a}\times\vec{b}=0\ \Rightarrow\ \big|\vec{a}\times\vec{b}\big|=0$ $\ \Rightarrow\ \ \big|\vec{a}\big|.\big|\vec{b}\big|\sin\theta=0$ $\therefore\ \ \big|\vec{a}\big|=0\ \ \text{or}\ \ \big|\vec{b}\big|=0\ \ \text{or}\ \ $ $\sin\theta=0\ \ \Rightarrow\ \theta=0^\circ$ $\Rightarrow\ \ \vec{a}=0\ \ \text{or}\ \ \vec{b}=0 \ \ \text{or}\ \ $ $\text{vector}\ \vec{a}\ \text{and}\ \vec{b}\ \text{are collinear or parallel.}\ \ \ \ \ ...\text{(ii)}$ Since, vectors $\vec{a}\ \&\ \vec{b}$ are perpendicular to each other as well as parallel are not possible. ...(iii) Therefore, form eq. (i), (ii) and (iii), $\ \text{either}\ \vec{a}=\vec{0}\ \ \ \text{or}\ \vec{b}=\vec{0}$ $\therefore\ \ \vec{a}.\vec{b}=0 \ \ \text{and}\ \ \vec{a}\times\vec{b}=0$
View full question & answer→Question 1492 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors, then write the value of $\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2.$
AnswerIt is given that $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors. $\Rightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=1\dots(1)$ Now, $\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2$ $=\big(|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\big)^2+\big(|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\big)^2$ $=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2(\cos^2\theta+\sin^2\theta)$ $=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$ (1) $=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$ $=1^21^2$ [From (1)]$=1$
View full question & answer→Question 1502 Marks
Find the position vector of a point R which divides the line segment joining points $\text{P}\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)$ and $\text{Q}\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$ in the ratio 2 : 1.Externally
AnswerGiven: R divides the line segment joining the points $\text{P}\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big),\text{Q}\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$ in the ratio 2 : 1 externally.
Therefore position vector of $\text{R}=\frac{2\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)+1\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)}{2+1}$
$=-3\hat{\text{i}}+\hat{\text{k}}$
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