MCQ 11 Mark
If $\theta$ is the angle between the vectors $2\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}$ and $3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}},$ then $\sin\theta=$
- A
$\frac{2}{3}$
- ✓
$\frac{2}{\sqrt{7}}$
- C
$\frac{\sqrt{2}}{7}$
- D
$\sqrt{\frac{2}{7}}$
AnswerCorrect option: B. $\frac{2}{\sqrt{7}}$
Let:
$\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{b}}=3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
$|\vec{\text{a}}|=\sqrt{2^2+(-2)^2+4^2}$
$=\sqrt{4+4+16}$
$=\sqrt{24}$
$=2\sqrt{6}$
$\Big|\vec{\text{b}}\big|=\sqrt{3^2+1^2+2^2}$
$=\sqrt{9+1+4}$
$=\sqrt{14}$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&-2&4\\3&1&2 \end{vmatrix}$
$=-8\hat{\text{i}}+8\hat{\text{j}}+8\hat{\text{k}}$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{64+64+64}$
$=\sqrt{192}$
$=8\sqrt{3}$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$\Rightarrow8\sqrt{3}=(2\sqrt{6})(\sqrt{14})\sin\theta$
$\Rightarrow\sin\theta=\frac{8\sqrt{3}}{4\sqrt{21}}$
$=\frac{2}{\sqrt{7}}$
$\Rightarrow\theta=\sin^{-1}\Big(\frac{2}{\sqrt{7}}\Big)$
View full question & answer→MCQ 21 Mark
A unit vector perpendicular to both $\hat{\text{i}}+\hat{\text{j}}$ and $\hat{\text{j}}+\hat{\text{k}}$ is:
- A
$\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
- B
$\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
- C
$\frac{1}{\sqrt{3}}\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
- ✓
$\frac{1}{\sqrt{3}}\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
AnswerCorrect option: D. $\frac{1}{\sqrt{3}}\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
Let:
$\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$
$\vec{\text{b}}=0\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
$\therefore\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&1&0\\0&1&1 \end{vmatrix}$
$=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{1+1+1}$
$=\sqrt{3}$
Unit vector perpendicular to $\vec{\text{a}}$ and $\vec{\text{b}}=\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\times\vec{\text{b}}\big|}=\frac{\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}}{\sqrt{3}}$
Disclaimer: The answer given for this question in the textbook is incorrect.
View full question & answer→MCQ 31 Mark
vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are inclined at angle $\theta=120^\circ.$ if $|\vec{\text{a}}|=1,\big|\vec{\text{b}}\big|=2,$ then $\big[\big(\vec{\text{a}}+3\vec{\text{b}}\big)\times\big(3\vec{\text{a}}-\vec{\text{b}}\big)\big]^2$ is equal to:
Answer$\big(\vec{\text{a}}+3\vec{\text{b}}\big)\times\big(3\vec{\text{a}}-\vec{\text{b}}\big)$
$=3\big(\vec{\text{a}}\times\vec{\text{a}}\big)-\vec{\text{a}}\times\vec{\text{b}}+9\big(\vec{\text{b}}\times\vec{\text{a}}\big)-3\big(\vec{\text{b}}\times\vec{\text{b}}\big)$
$=3(0)-\vec{\text{a}}\times\vec{\text{b}}-9\big(\vec{\text{a}}\times\vec{\text{b}}\big)-3(0)$
$=-10\big(\vec{\text{a}}\times\vec{\text{b}}\big)$
Now,
$\big|\big(\vec{\text{a}}\times3\vec{\text{b}}\big)\times\big(3\vec{\text{a}}-\vec{\text{b}}\big)\big|^2$
$=\big|-10\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|^2$
$=100\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|^2$
$=100|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2\sin^2120$
$=100(1)^2(2)^2\Big(\frac{\sqrt{3}}{2}\Big)^2$
$=400\times\frac{3}{4}$
$=300$
View full question & answer→MCQ 41 Mark
If $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=4,\big|\vec{\text{a}}.\vec{\text{b}}\big|=2,$ then $|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2=$
AnswerWe know
$\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+\big|\vec{\text{a}}\times\vec{\text{b}}\big|62=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2\dots(1)$
$\big|\vec{\text{a}}.\vec{\text{b}}\big|=2$ (Given)
$\Rightarrow\big|\vec{\text{a}}.\vec{\text{b}}\big|^2=\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
From $(1),$ we get
$(2)^2+(4)^2=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$
$\Rightarrow|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2=20$
View full question & answer→MCQ 51 Mark
The value of $\hat{\text{i}}.\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\hat{\text{j}}.\big(\hat{\text{i}}\times\hat{\text{k}}\big)+\hat{\text{k}}.\big(\hat{\text{i}}\times\hat{\text{j}}\big),$ is:
Answer$\hat{\text{i}}.\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\hat{\text{j}}.\big(\hat{\text{i}}\times\hat{\text{k}}\big)+\hat{\text{k}}.\big(\hat{\text{i}}\times\hat{\text{j}}\big)$
$=\hat{\text{i}}.\hat{\text{i}}+\hat{\text{j}}.(-\hat{\text{j}})+\hat{\text{k}}.\hat{\text{k}}$
$=|\hat{\text{i}}|^2-|\hat{\text{j}}|^2+|\hat{\text{k}}|^2$
$=1-1+1$
$=1$
View full question & answer→MCQ 61 Mark
If $\vec{\text{a}}$ is any vector, then $\big(\vec{\text{a}}\times\hat{\text{i}}\big)^2+\big(\vec{\text{a}}\times\hat{\text{j}}\big)^2+\big(\vec{\text{a}}\times\hat{\text{k}}\big)^2=$
- A
$\vec{\text{a}}^2$
- ✓
$2\vec{\text{a}}^2$
- C
$3\vec{\text{a}}^2$
- D
$4\vec{\text{a}}^2$
AnswerCorrect option: B. $2\vec{\text{a}}^2$
Let $\vec{\text{a}}={\text{a}}_1\hat{\text{i}}+{\text{a}}_2\hat{\text{j}}+{\text{a}}_3\hat{\text{k}}$
$\vec{\text{a}}\times\hat{\text{i}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{a}_2&\text{a}_3\\1&0&0 \end{vmatrix}$
$=\text{a}_3\hat{\text{j}}-\text{a}_2\hat{\text{k}}$
$\Rightarrow\big(\vec{\text{a}}\times\hat{\text{i}}\big)^2=\big(\text{a}_3\hat{\text{j}}-\text{a}_2\hat{\text{k}}\big)^2$
$={\text{a}_3}^2|\hat{\text{j}}|^2+{\text{a}_2}^2|\hat{\text{k}}|^2-2\text{a}_3\text{a}_2\big(\hat{\text{j}}.\hat{\text{k}}\big)$
$={\text{a}_3}^2+{\text{a}_2}^2$ $\big(\because\hat{\text{j}}.\hat{\text{k}}=0\dots(1)\big)$
$\therefore\vec{\text{a}}\times\hat{\text{j}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{a}_2&\text{a}_3\\0&1&0 \end{vmatrix}$
$=-\text{a}_3\hat{\text{i}}+\text{a}_1\hat{\text{k}}$
$\Rightarrow\big(\vec{\text{a}}\times\hat{\text{j}}\big)^2=\big(-\text{a}_3\hat{\text{i}}+\text{a}_1\hat{\text{k}}\big)^2$
$={\text{a}_3}^2|\hat{\text{i}}|^2+{\text{a}_1}^2|\hat{\text{k}}|^2-2\text{a}_3\text{a}_2\big(\hat{\text{i}}.\hat{\text{k}}\big)$
$={\text{a}_3}^2+{\text{a}_1}^2$ $(\because\hat{\text{i}}.\hat{\text{k}}=0)\dots(2)$
$\therefore\vec{\text{a}}\times\hat{\text{k}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{a}_2&\text{a}_3\\0&0&1 \end{vmatrix}$
$=\text{a}_2\hat{\text{i}}-\text{a}_1\hat{\text{j}}$
$\Rightarrow\big(\vec{\text{a}}\times\hat{\text{k}}\big)^2=\big(\text{a}_2\hat{\text{i}}-\text{a}_1\hat{\text{j}}\big)^2$
$={\text{a}_2}^2|\hat{\text{i}}|^2+{\text{a}_1}^2|\hat{\text{j}}|^2-2\text{a}_1\text{a}_2\big(\hat{\text{i}}.\hat{\text{j}}\big)$
$={\text{a}_2}^2+{\text{a}_1}^2$ $(\because\hat{\text{i}}.\hat{\text{j}}=0)\dots(3)$
Adding $(1), (2)$ and $(3),$ we get
$\big(\vec{\text{a}}\times\hat{\text{i}}\big)^2+\big(\vec{\text{a}}\times\hat{\text{j}}\big)^2+\big(\vec{\text{a}}\times\hat{\text{k}}\big)^2={\text{a}_3}^2+{\text{a}_2}^2+{\text{a}_3}^2+{\text{a}_1}^2+{\text{a}_2}^2+{\text{a}_1}^2$
$=2\big({\text{a}_1}^2+{\text{a}_2}^2+{\text{a}_3}^2\big)$
$=2\vec{\text{a}}^2$ $\big(\because|\vec{\text{a}}|=\sqrt{{\text{a}_1}^2+{\text{a}_2}^2+{\text{a}_3}^2}\big)$
View full question & answer→MCQ 71 Mark
If $\theta$ is the angle between any two vectors $\vec{\text{a}}$ and $\vec{\text{b}},$ then $\big|\vec{\text{a}}.\vec{\text{b}}\big|=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$ when $\theta$ is equal to:
- A
$0$
- ✓
$\frac{\pi}{4}$
- C
$\frac{\pi}{2}$
- D
$\pi$
AnswerCorrect option: B. $\frac{\pi}{4}$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
We know
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow\big|\vec{\text{a}}.\vec{\text{b}}\big|=\big||\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big||\cos\theta|$
Given: $\big|\vec{\text{a}}.\vec{\text{b}}\big|=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big||\cos\theta|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$\Rightarrow|\cos\theta|=\sin\theta$
$\Rightarrow\theta=\frac{\pi}{4}$
View full question & answer→MCQ 81 Mark
If $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{c}}=-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}},$ then a unit vector normal to the vectors $\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{b}}-\vec{\text{c}}$ is:
- ✓
$\hat{\text{i}}$
- B
$\hat{\text{j}}$
- C
$\hat{\text{k}}$
- D
$\text{None of these}$
AnswerCorrect option: A. $\hat{\text{i}}$
$\vec{\text{a}}+\vec{\text{b}}=0\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}-\vec{\text{c}}=0\hat{\text{i}}-0\hat{\text{j}}+3\hat{\text{k}}$
$\big(\vec{\text{a}}\times\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\0&3&1\\0&0&3 \end{vmatrix}$
$=9\hat{\text{i}}$
$\big|\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|=9|\hat{\text{i}}|$
$=9(1)$
$=9$
Unit vector perpendicular to both $\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{b}}-\vec{\text{c}}=\frac{\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)}{\big|\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|}$
$=\frac{9\hat{\text{i}}}{9}$
$=\hat{\text{i}}$
View full question & answer→MCQ 91 Mark
The vector $\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{k}}$ is to be written as the sum of a vector $\vec{\alpha}$ parallel to $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}$ and a vector $\vec{\beta}$ perpendicular to $\vec{\text{a}}.$ Then $\vec{\alpha}=$
- ✓
$\frac{3}{2}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
- B
$\frac{2}{3}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
- C
$\frac{1}{2}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
- D
$\frac{1}{3}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
AnswerCorrect option: A. $\frac{3}{2}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
Let:
$\vec{\alpha}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$\vec{\beta}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$
Now,
$\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{k}}=\vec{\alpha}+\vec{\beta} ($Given$)$
$\Rightarrow3\hat{\text{i}}+0\hat{\text{j}}+4\hat{\text{k}}=(\text{a}_1+\text{b}_1)\hat{\text{i}}+(\text{a}_2+\text{b}_2)\hat{\text{j}}+(\text{a}_3+\text{b}_3)\hat{\text{k}}$
$\Rightarrow\text{a}_1+\text{b}_1=3;\text{a}_2+\text{b}_2=0;\text{a}_3+\text{b}_3=4$
$\Rightarrow\text{a}_1+\text{b}_1=3;\text{a}_2=-\text{b}_2;\text{a}_3+\text{b}_3=4\dots(1)$
$\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}} ($Given$)$
Also, $\vec{\alpha}$ is parallrl to $\vec{\text{a}}.$
$\Rightarrow\vec{\alpha}\times\vec{\text{a}}=\vec{0}$
$\Rightarrow\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{a}_2&\text{a}_3\\1&1&0\end{vmatrix}=\vec{0}$
$\Rightarrow-\text{a}_3\hat{\text{i}}+\text{a}_3\hat{\text{j}}+(\text{a}_1-\text{a}_2)\hat{\text{k}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$
$\Rightarrow\text{a}_3=0;\text{a}_1-\text{a}_2=0$
$\Rightarrow\text{a}_3=0;\text{a}_1=\text{a}_2\dots(2)$
Since $\vec{\beta}$ is perpendicular to $\vec{\text{a}},$ we get
$\Rightarrow\vec{\beta}.\vec{\text{a}}=0$
$\Rightarrow\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big).\big(\hat{\text{i}}.\hat{\text{j}}\big)=0$
$\Rightarrow\text{b}_1+\text{b}_2=0$
$\Rightarrow\text{b}_1=-\text{b}_2\dots(3)$
Solving $(1), (2)$ and $(3)$, we get
$\text{a}_1=\frac{3}{2};\text{a}_2=\frac{3}{2};\text{a}_3=0$
$\therefore\vec{\alpha}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$=\frac{3}{2}\hat{\text{i}}+\frac{3}{2}\hat{\text{j}}+0\hat{\text{k}}$
$=\frac{3}{2}\big(\hat{\text{i}}+\hat{\text{j}}\big)$
View full question & answer→MCQ 101 Mark
If $\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}},$ then $\vec{\text{a}}\times\vec{\text{b}}$ is:
- A
$10\hat{\text{i}}+2\hat{\text{j}}+11\hat{\text{k}}$
- ✓
$10\hat{\text{i}}+3\hat{\text{j}}+11\hat{\text{k}}$
- C
$10\hat{\text{i}}-3\hat{\text{j}}+11\hat{\text{k}}$
- D
$10\hat{\text{i}}-2\hat{\text{j}}-10\hat{\text{k}}$
AnswerCorrect option: B. $10\hat{\text{i}}+3\hat{\text{j}}+11\hat{\text{k}}$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&-3&-1\\1&4&-2 \end{vmatrix}$
$=10\hat{\text{i}}+3\hat{\text{j}}+11\hat{\text{k}}$
View full question & answer→MCQ 111 Mark
The unit vector perpendicular to the plane passing through points $\text{P}\big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big),\text{Q}\big(2\hat{\text{i}}-\hat{\text{k}}\big)$ and $\text{R}\big(2\hat{\text{j}}+\hat{\text{k}}\big)$ is:
- A
$2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
- B
$\sqrt{6}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
- ✓
$\frac{1}{\sqrt{6}}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
- D
$\frac{1}{6}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
AnswerCorrect option: C. $\frac{1}{\sqrt{6}}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
The vector $\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}$ is perpendicular to the vectors $\overrightarrow{\text{PQ}}$ and $\overrightarrow{\text{PR}}.$
$\therefore$ Required unit vector $=\frac{\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}}{\big|\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}\big|}$
Now,
$\overrightarrow{\text{PQ}}=\text{P.V}\text{ of }\text{Q}-\text{P.V}.\text{ of P}$
$=\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}$
$\overrightarrow{\text{PR}}=\text{P.V}\text{ of }\text{R}-\text{P.V}.\text{ of P}$
$=-\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$
$\therefore\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&1&-3\\-1&3&-1 \end{vmatrix}$
$=8\hat{\text{i}}+4\hat{\text{j}}+4\hat{\text{k}}$
$=4\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
$\Rightarrow\big|\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}\big|=\sqrt{64+16+16}$
$=\sqrt{96}$
$=4\sqrt{6}$
Required unit vector $=\frac{\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}}{\big|\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}\big|}$
$=\frac{4\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)}{4\sqrt{6}}$
$=\frac{1}{\sqrt{6}}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
View full question & answer→MCQ 121 Mark
If $\vec{\text{a}}.\vec{\text{b}}=\vec{\text{a}}.\vec{\text{c}}$ and $\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{a}}\times\vec{\text{c}}.\vec{\text{a}}\neq0,$ then:
AnswerCorrect option: A. $\vec{\text{b}}=\vec{\text{c}}$
$\vec{\text{a}}.\vec{\text{b}}=\vec{\text{a}}.\vec{\text{c}}$
$\Rightarrow\vec{\text{a}}.\vec{\text{b}}-\vec{\text{a}}.\vec{\text{c}}=0$
$\Rightarrow\vec{\text{a}}.\big(\vec{\text{b}}-\vec{\text{c}}\big)=0$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\big(\vec{\text{b}}-\vec{\text{c}}\big)$
$|\vec{\text{a}}|\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\cos\theta\dots(1)$
and $\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{a}}\times\vec{\text{c}}$
$\Rightarrow\vec{\text{a}}\times\vec{\text{b}}-\vec{\text{a}}\times\vec{\text{c}}=0$
$\Rightarrow\vec{\text{a}}\times\big(\vec{\text{b}}-\vec{\text{c}}\big)=0$
Then, $|\vec{\text{a}}|\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\sin\theta=0\dots(2)$
Here, it is given that $\vec{\text{a}}\neq0$
Therefore, for eq. $(1)$ and eq. $(2)$ to be $0$
We have,
$\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\cos\theta=0$
For $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\cos\theta=0,$ one of $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|$ or $\cos\theta$ must be $0$
Case 1 :
Let $\cos\theta=0$
$\Rightarrow\theta=90^\circ$
$\Rightarrow\sin\theta=1$
$\&$ if $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\sin\theta=0$ and $\sin\theta=1$
Then $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|=0$
$\Rightarrow\vec{\text{b}}=\vec{\text{c}}$
Case 2 :
Let $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|=0$
$\Rightarrow\vec{\text{b}}=\vec{\text{c}}$
Hence, $\vec{\text{b}}=\vec{\text{c}}$
View full question & answer→MCQ 131 Mark
The value of $\big(\vec{\text{a}}\times\vec{\text{b}}\big)^2$ is:
- A
$|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
- ✓
$|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2-\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
- C
$|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\big(\vec{\text{a}}.\vec{\text{b}}\big)$
- D
$|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-\vec{\text{a}}.\vec{\text{b}}$
AnswerCorrect option: B. $|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2-\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
$\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2$
$=\big(|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\big)^2+\big(|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\big)^2$
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2(\cos^2\theta+\sin^2\theta)$
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2 (1)$
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$
$\therefore\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2-\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
Thus, the value of $\big(\vec{\text{a}}\times\vec{\text{b}}\big)^2$ is $|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2-\big(\vec{\text{a}}.\vec{\text{b}}\big)^2.$
View full question & answer→MCQ 141 Mark
If $\vec{\text{a}},\vec{\text{b}}$ represent the diagonals of a rhombus, then:
- A
$\vec{\text{a}}\times\vec{\text{b}}=\vec{0}$
- ✓
$\vec{\text{a}}.\vec{\text{b}}=0$
- C
$\vec{\text{a}}.\vec{\text{b}}=1$
- D
$\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{a}}$
AnswerCorrect option: B. $\vec{\text{a}}.\vec{\text{b}}=0$
We know that the diagonals in a rhombus $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular.
Therefore, their dot product is zero.
$\Rightarrow\vec{\text{a}}.\vec{\text{b}}=0$
View full question & answer→MCQ 151 Mark
If $\hat{\text{i}},\hat{\text{j}},\hat{\text{k}}$ are unit vectors, then
- A
$\hat{\text{i}}.\hat{\text{j}}=1$
- ✓
$\hat{\text{i}}.\hat{\text{i}}=1$
- C
$\hat{\text{i}}\times\hat{\text{j}}=1$
- D
$\hat{\text{i}}\times\big(\hat{\text{j}}\times\hat{\text{k}}\big)=1$
AnswerCorrect option: B. $\hat{\text{i}}.\hat{\text{i}}=1$
Let us check each option one by one.
$a.$ We know
$\hat{\text{i}}.\hat{\text{j}}=0$
$\neq1$
$b.$ We know
$\hat{\text{i}}.\hat{\text{i}}=|\hat{\text{i}}|^2$
$=1^2$
$=1$
$c. \hat{\text{i}}\times\hat{\text{j}}=\hat{\text{k}}$
$\neq1$
$d. \hat{\text{i}}\times\big(\hat{\text{i}}\times\hat{\text{k}}\big)=\hat{\text{i}}\times\hat{\text{i}}$
$=0$
$\neq1$
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