Question 12 Marks
For any two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ write the value of $\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2$ in terms of their magnitudes.
Answer$\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2$
$=\big(|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\big)^2+\big(|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\big)^2$
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$ $\big(\cos^2\theta+\sin^2\theta\big)$
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$ (1)
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$
View full question & answer→Question 22 Marks
For any two vectors $\vec{\text{a}}$ and $\vec{\text{b}},$ find $\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{b}}.$
AnswerLet:$\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$\vec{\text{b}}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\end{vmatrix}$
$=\hat{\text{i}}(\text{a}_2\text{b}_3-\text{a}_3\text{b}_2)-\hat{\text{j}}(\text{a}_1\text{b}_3-\text{a}_3\text{b}_1)+\hat{\text{k}}(\text{a}_1\text{b}_2-\text{a}_2\text{b}_1)$
$\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{b}}$
$=\big[\hat{\text{i}}(\text{a}_2\text{b}_3-\text{a}_3\text{b}_2)-\hat{\text{j}}(\text{a}_1\text{b}_3-\text{a}_3\text{b}_1)+\hat{\text{k}}(\text{a}_1\text{b}_2-\text{a}_2\text{b}_1)\big]\\.\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
$=\text{a}_2\text{b}_1\text{b}_3-\text{a}_3\text{b}_1\text{b}_2-\text{a}_1\text{b}_2\text{b}_3+\text{a}_3\text{b}_1\text{b}_2+\text{a}_1\text{b}_2\text{b}_3-\text{a}_2\text{b}_1\text{b}_3$
$=\text{b}_1(\text{a}_2\text{b}_3-\text{a}_3\text{b}_2)-\text{b}_2(\text{a}_1\text{b}_3-\text{a}_3\text{b}_1)+\text{b}_3(\text{a}_1\text{b}_2-\text{a}_2\text{b}_1)$
$=\text{a}_2\text{b}_1\text{b}_3-\text{a}_3\text{b}_1\text{b}_2-\text{a}_1\text{b}_2\text{b}_3+\text{a}_3\text{b}_1\text{b}_2+\text{a}_1\text{b}_2\text{b}_3-\text{a}_2\text{b}_1\text{b}_3$
$=0$
View full question & answer→Question 32 Marks
If $\vec{\text{a}}=3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},$ then find $\big(\vec{\text{a}}\times\vec{\text{b}}\big)\vec{\text{a}}.$
AnswerSince $\vec{\text{a}}\times\vec{\text{b}}$ is a vector, $\big(\vec{\text{a}}\times\vec{\text{b}}\big)\vec{\text{a}}$ without any dot or cross product in between is meaningless.
View full question & answer→Question 42 Marks
If $\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2=144$ and $|\vec{\text{a}}|=4,$ find $\big|\vec{\text{b}}\big|.$
AnswerWe know
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$
$\Rightarrow144=4^2\big|\vec{\text{b}}\big|^2$
$\Rightarrow144=16\big|\vec{\text{b}}\big|^2$
$\Rightarrow\big|\vec{\text{b}}\big|^2=9$
$\Rightarrow\big|\vec{\text{b}}\big|=3$
View full question & answer→Question 52 Marks
Write the value of $\hat{\text{i}}.\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\hat{\text{j}}.\big(\hat{\text{k}}\times\hat{\text{i}}\big)+\hat{\text{k}}.\big(\hat{\text{i}}\times\hat{\text{j}}\big).$
Answer$\hat{\text{i}}.\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\hat{\text{j}}.\big(\hat{\text{k}}\times\hat{\text{i}}\big)+\hat{\text{k}}.\big(\hat{\text{j}}\times\hat{\text{i}}\big)$
$=\hat{\text{i}}.\hat{\text{i}}+\hat{\text{j}}.\hat{\text{j}}+\hat{\text{k}}.\hat{\text{k}}$
$=|\hat{\text{i}}|^2+|\hat{\text{j}}|^2+|\hat{\text{k}}|^2$
$=1+1+1$
$=3$
View full question & answer→Question 62 Marks
vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are such that $|\vec{\text{a}}|=\sqrt{3},\big|\vec{\text{b}}\big|=\frac{2}{3}$ and $\big(\vec{\text{a}}\times\vec{\text{b}}\big)$ is a unit vector. write the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
AnswerGiven: $\vec{\text{a}}\times\vec{\text{b}}$ is a unit vector.
$\Rightarrow\big|\vec{\text{a}}\times\vec{\text{b}}\big|=1\dots(1)$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
We know
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
From (1), we get
$1=(\sqrt{3})\Big(\frac{2}{3}\Big)\sin\theta$
$\Rightarrow\sin\theta=\frac{\sqrt{3}}{2}$
$\Rightarrow\theta=\frac{\pi}{3}$
View full question & answer→Question 72 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors such that $\vec{\text{a}}\times\vec{\text{b}}$ is also a unit vector, find the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
Given:
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=1$
$|\vec{\text{a}}|=1$
$|\vec{\text{b}}|=1$
We know
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$\Rightarrow1=(1)(1)\sin\theta$
$\Rightarrow\sin\theta=1$
$\Rightarrow\theta=\frac{\pi}{2}$
View full question & answer→Question 82 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors of magnitudes 3 and $\frac{\sqrt{2}}{3}$ repectively such that $\vec{\text{a}}\times\vec{\text{b}}$ is a unit vector. write the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
AnswerWrite the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
It is given that $\vec{\text{a}}\times\vec{\text{b}}$ is a unit vector.
$\Rightarrow\big|\vec{\text{a}}\times\vec{\text{b}}\big|=1$
We know
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$\Rightarrow1=(3)\Big(\frac{\sqrt{2}}{3}\big)\sin\theta$
$\Rightarrow\sin\theta=\frac{1}{\sqrt{2}}$
$\Rightarrow\theta=45^\circ,135^\circ$
View full question & answer→Question 92 Marks
If $\vec{\text{a}}=\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}}$ and $\vec{\text{b}}=-\hat{\text{i}}+3\hat{\text{k},}$ find $\big|\vec{\text{a}}\times\vec{\text{b}}\big|.$
AnswerIf $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{b}_1\hat{\text{j}}+\text{c}_1\hat{\text{k}}$ and
$\vec{\text{b}}=\text{a}_2\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{c}_2\hat{\text{k}},$ then
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{b}_1&\text{c}_2\\\text{a}_2&\text{b}_2&\text{c}_2 \end{vmatrix}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&3&-2\\-1&0&3 \end{vmatrix}$
$=\hat{\text{i}}(9-0)-\hat{\text{j}}(3-2)+\hat{\text{k}}(0+3)$
$\vec{\text{a}}\times\vec{\text{b}}=9\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{(9)^2+(-1)^2+(3)^2}$
$=\sqrt{81+1+9}$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{91}$
View full question & answer→Question 102 Marks
Write the expression for the area of the parallelogram having $\vec{\text{a}}$ and $\vec{\text{b}}$ as its diagonals.
AnswerGiven: $\vec{\text{a}}$ and $\vec{\text{b}}$ are diagonals of a parallelogram.
Area of the parallelogram $=\frac{1}{2}\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
View full question & answer→Question 112 Marks
For any three vectors $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ write the value of $\vec{\text{a}}\times\big(\vec{\text{b}}+\vec{\text{c}}\big)+\vec{\text{b}}\times\big(\vec{\text{c}}+\vec{\text{a}}\big)+\vec{\text{c}}\times\big(\vec{\text{a}}+\vec{\text{b}}\big).$
Answer$\vec{\text{a}}\times\big(\vec{\text{b}}+\vec{\text{c}}\big)+\vec{\text{b}}\times\big(\vec{\text{c}}+\vec{\text{a}}\big)+\vec{\text{c}}\times\big(\vec{\text{a}}+\vec{\text{b}}\big)$
$=\big(\vec{\text{a}}\times\vec{\text{b}}\big)+\big(\vec{\text{a}}\times\vec{\text{c}}\big)+\big(\vec{\text{b}}\times\vec{\text{c}}\big)+\big(\vec{\text{b}}\times\vec{\text{a}}\big)+\big(\vec{\text{c}}\times\vec{\text{a}}\big)+\big(\vec{\text{c}}\times\vec{\text{b}}\big)$
$=\big(\vec{\text{a}}\times\vec{\text{b}}\big)+\big(\vec{\text{a}}\times\vec{\text{c}}\big)+\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\big(\vec{\text{a}}\times\vec{\text{b}}\big)-\big(\vec{\text{a}}\times\vec{\text{c}}\big)-\big(\vec{\text{b}}\times\vec{\text{c}}\big)$
$=\vec{0}$
View full question & answer→Question 122 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors such that $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{3}$ and $\vec{\text{a}}.\vec{\text{b}}=1,$ find the angle between.
Answer$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{3}$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta=\sqrt{3}\dots(1)$
$\vec{\text{a}}.\vec{\text{b}}=1$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=1\dots(2)$
Dividing (1) by (2), we get
$\frac{|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta}=\sqrt{3}$
$\Rightarrow\tan\theta=\sqrt{3}$
$\Rightarrow\theta=60^\circ$
View full question & answer→Question 132 Marks
Write a unit vector perpendicular to $\hat{\text{i}}+\hat{\text{j}}$ and $\hat{\text{j}}+\hat{\text{k}}.$
AnswerLet $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}};\vec{\text{b}}=0\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&1&0\\0&1&1 \end{vmatrix}$
$=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{1+1+1}$
$=\sqrt{3}$
Unit vector perpendicular to $\vec{\text{a}}$ and $\vec{\text{b}}$ is, $\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\times\vec{\text{b}}\big|}=\frac{1}{\sqrt{3}}\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
View full question & answer→Question 142 Marks
For any two vectors $\vec{\text{a}}$ and $\vec{\text{b}},$ find $\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{a}}\big).$
AnswerLet:
$\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$\vec{\text{b}}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$
$\vec{\text{b}}\times\vec{\text{a}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{a}_1&\text{a}_2&\text{a}_3\end{vmatrix}$
$=\hat{\text{i}}(\text{b}_2\text{a}_3-\text{b}_3\text{a}_2)-\hat{\text{j}}(\text{b}_1\text{a}_3-\text{b}_3\text{a}_1)+\hat{\text{k}}(\text{b}_1\text{a}_2-\text{b}_2\text{a}_1)$
Now,
$\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{a}}\big)$
$=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big).\big[\hat{\text{i}}(\text{b}_2\text{a}_3-\text{b}_3\text{a}_2)\\-\hat{\text{j}}(\text{b}_1\text{a}_3-\text{b}_3\text{a}_1)+\hat{\text{k}}(\text{b}_1\text{a}_2-\text{b}_2\text{a}_1)\big]$
$=\text{a}_1(\text{b}_2\text{a}_3-\text{b}_3\text{a}_2)-\text{a}_2(\text{b}_1\text{a}_3-\text{b}_3\text{a}_1)+\text{a}_3(\text{b}_1\text{a}_2-\text{b}_2\text{a}_1)$
$=\text{a}_1\text{b}_2\text{a}_3-\text{a}_1\text{b}_3\text{a}_2-\text{a}_2\text{b}_1\text{a}_3+\text{a}_2\text{b}_3\text{a}_1+\text{a}_3\text{b}_1\text{a}_2-\text{a}_3\text{b}_2\text{a}_1$
$=0$
View full question & answer→Question 152 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 162 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 172 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 182 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 192 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 202 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 212 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 222 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 232 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 242 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 252 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 262 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 272 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 282 Marks
If $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}},$ then write the value of $\big|\vec{\text{r}}\times\hat{\text{i}}\big|^2.$
AnswerGiven: $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
Now,
$\vec{\text{i}}=\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$
$\vec{\text{r}}\times\vec{\text{i}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{x}&\text{y}&\text{z}\\1&0&0 \end{vmatrix}$
$=0\hat{\text{i}}+\text{z}\hat{\text{j}}-\text{y}\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{r}}\times\vec{\text{i}}\big|=\sqrt{\text{x}^2+\text{y}^2}$
$\Rightarrow\big|\vec{\text{r}}\times\vec{\text{i}}\big|^2=\text{x}^2+\text{y}^2$
View full question & answer→Question 292 Marks
Find $\lambda,$ if $\big(2\hat{\text{i}}+6\hat{\text{j}}+14\hat{\text{k}}\big)\times\big(\hat{\text{i}}-\lambda\hat{\text{j}}+7\hat{\text{k}}\big)=\vec{0}.$
AnswerGiven: $\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&6&14\\1&-\lambda&7\end{vmatrix}=\vec{0}$
$\Rightarrow\hat{\text{i}}(42+14\lambda)-0\hat{\text{j}}+\hat{\text{k}}(-2\lambda-6)=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$
$\Rightarrow42+14\lambda=0;-2\lambda-6=0$
$\Rightarrow\lambda=-3$ (This satisfies the above equations)
View full question & answer→Question 302 Marks
Write the value of the area of the parallelogram determined by the vectors $2\hat{\text{i}}$ and $3\hat{\text{j}}.$
AnswerLet:
$\vec{\text{a}}=2\hat{\text{i}}$
$\vec{\text{b}}=3\hat{\text{j}}$
$\vec{\text{a}}\times\vec{\text{b}}=6\big(\hat{\text{i}}\times\hat{\text{j}}\big)$
$=6\hat{\text{k}}$
Area of the parallelogram $=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$=6|\hat{\text{k}}|$
$=6(1)$
$=6\text{ sq. units}$
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Find the angle between two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ with magnitudes 1 and 2 respectively and when $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{3.}$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
We know $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$\Rightarrow\sqrt{3}=(1)(2)\sin\theta$
$\Rightarrow\sin\theta=\frac{\sqrt{3}}{2}$
$\Rightarrow\theta=\frac{\pi}{3}$
View full question & answer→Question 322 Marks
If $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},$ find the magnitude of $\vec{\text{a}}\times\vec{\text{b}}.$
AnswerGiven:$\vec{\text{a}}=2\hat{\text{i}}+0\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&0&1\\1&1&1 \end{vmatrix}$
$=(0-1)\hat{\text{i}}-(2-1)\hat{\text{j}}+(2-0)\hat{\text{k}}$
$=-\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{(-1)^2+(1-)^2+2^2}$
$=\sqrt{6}$
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