Question 1012 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 1022 Marks
Find the angle between the vectora $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}.$
AnswerWe have
$\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
$|\vec{\text{a}}|=\sqrt{(1)^2+(-1)^2+(1)^2}=\sqrt{3}$
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2+(1)^2+(-1)^2}=\sqrt{3}$
and
$\vec{\text{a}}.\vec{\text{b}}=1-1-1=-1$
Now,
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|}=\frac{-1}{\sqrt{3}\sqrt{3}}=\frac{-1}{3}$
$\therefore\theta=\cos^-1\big(\frac{-1}{3}\big)$
View full question & answer→Question 1032 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are position vectors of the vertices A, B and C respectively, of a triangle ABC, write the value of$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}$.
AnswerGiven: $\vec{\text{a}},\vec{\text{b}}\text{ and }\vec{\text{c}}$ are the position vectors of A, B and C respectively. Then,
$\overrightarrow{\text{AB}}=\vec{\text{b}}-\vec{\text{a}}$
$\overrightarrow{\text{BC}}=\vec{\text{c}}-\vec{\text{b}}$
$\overrightarrow{\text{CA}}=\vec{\text{a}}-\vec{\text{c}}$
Consider,
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}\\=\vec{\text{b}}-\vec{\text{a}}+\vec{\text{c}}-\vec{\text{b}}+\vec{\text{a}}-\vec{\text{c}}$
$=\vec0$
View full question & answer→Question 1042 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 1052 Marks
Find the unit vector parallel to the vector $\hat{\text{i}}+\sqrt3\hat{\text{j}}$.
AnswerLet $\vec{\text{a}}=\hat{\text{i}}+\sqrt3\hat{\text{j}}$
Then, $\big|\vec{\text{a}}\big|=\sqrt{1^2+\big(\sqrt3\big)^2}$
$=\sqrt{1+3}$
$=\sqrt4$
$=2$
Unit vector parallel to $\vec{\text{a}}=\hat{\text{a}}=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{1}{2}\big(\hat{\text{i}}+\sqrt3\hat{\text{j}}\big)=\frac{1}2\hat{\text{i}}+\frac{\sqrt3}2\hat{\text{j}}$
View full question & answer→Question 1062 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.
AnswerWe have, $\text{x}\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$ is a unit vector.
$\therefore\ \sqrt{\text{x}^2+\text{x}^2+\text{x}^2}=1$
$\Rightarrow\sqrt3|\text{x}|=1$
$\Rightarrow|\text{x}|=\frac{1}{\sqrt3}$
$\Rightarrow\text{x}=\pm\frac{1}{\sqrt3}$
View full question & answer→Question 1072 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 1082 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 1092 Marks
Find a unit vector in the direction of $\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}$.
AnswerGiven:
$\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}$
$|\vec{\text{a}}|=\sqrt{2^2+(-3)^2+6^2}$
$=\sqrt{4+9+36}$
$=\sqrt{49}$
$=7$
Unit vector $=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}}{7}$
$=\frac{2}7\hat{\text{i}}-\frac{3}7\hat{\text{j}}+\frac{6}7\hat{\text{k}}$
View full question & answer→Question 1102 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 1112 Marks
Write the value of $\lambda$ so that vectora $\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$ are perpendicular to each other.
AnswerWe have
$\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
The given vectors are perpendicular. so, their dot product is zero.
$\big(2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}\big).\big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big)$
$\Rightarrow2-2\lambda+3=0$
$\Rightarrow5-2\lambda=0$
$\Rightarrow-2\lambda=-5$
$\Rightarrow\lambda=\frac{5}{2}$
View full question & answer→Question 1122 Marks
Write the direction cosines of the vector $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ .
AnswerGiven: $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
Then, direction cosines are
$\frac{1}{\sqrt{1^2+2^2+3^2}},\frac{2}{\sqrt{1^2+2^2+3^2}},\frac{3}{\sqrt{1^2+2^2+3^2}}$ or, $\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}$
View full question & answer→Question 1132 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 1142 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 1152 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 1162 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 1172 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 1182 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 $\text{R}=\frac{2\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)+1\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)}{2+1}$
$=-3\hat{\text{i}}+\hat{\text{k}}$
View full question & answer→Question 1192 Marks
Find the angle between the vectors $\vec{\text{a}} $ and $\vec{\text{b}},$ where$\vec{\text{a}}=3\hat {\text{i}}-2\hat{\text{j}}-6\hat{\text{k}}$ and $\vec{\text{b}} =4\hat{\text{i}}-\hat{\text{j}}+8\hat{\text{k}}$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.
$\big|\vec{\text{a}}\big|=\sqrt{(3)^2+(-2)^2+(-6)^{2}}=\sqrt{49}=7$
$\big|\vec{\text{b}}\big|=\sqrt{(4)^2+(1)^2+(8)^{2}}=\sqrt{81}=9$
$\vec{\text{a}}.\vec{\text{b}}=12+2-48=-34$
$\cos\theta=\frac{\vec{\text{a }}.\vec{\text{ b}}}{\big|\vec{a}\big|\big|\vec{b}\big|}=\frac{-34}{(7)(9)}=\frac{-34}{63}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{-34}{63}\big)$
View full question & answer→Question 1202 Marks
A vector $\vec{\text{r}}$ is inclined at equal acute angles to x-axis, y-axis and z-axis. If $|\vec{\text{r}}|=6$ units, find $\vec{\text{r}}$.
AnswerHere, $\alpha=\beta=\gamma$
$\Rightarrow\ \cos\alpha=\cos\beta=\cos\gamma$
$\Rightarrow\ \text{l}=\text{m}=\text{n}=\text{x}$ (say)
We know that,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\text{x}^2+\text{x}^2+\text{x}^2=1$
$3\text{x}^2=1$
$\text{x}^2=\frac{1}3$
$\text{x}=\pm\frac{1}{\sqrt3}$
$\text{l}=\pm\frac{1}{\sqrt3},\ \text{m}=\pm\frac{1}{\sqrt3},\ \text{n}=\pm\frac{1}{\sqrt3}$
View full question & answer→Question 1212 Marks
Write the projection of the vector $\hat{\text{i}}+3\hat{\text{j}}+7\hat{\text{k}}$ on the vector $2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}.$
AnswerWe know that projection of $\vec{\text{a}}$ on $\vec{\text{b}}=\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}.$
Let $\vec{\text{a}}=\hat{\text{i}}+3\hat{\text{j}}+7\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}.$
$\therefore$ projection of $\vec{\text{a}}$ on $\vec{\text{b}}$
$=\frac{\big(\hat{\text{i}}+3\hat{\text{j}}+7\vec{\text{k}}\big).\big(2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}.\big)}{\big|2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}.\big|}$
$=\frac{1\times2+3\times(-3)+7\times6}{\sqrt{2^2+(-3)^2+6^2}}$
$=\frac{2-9+42}{\sqrt{49}}$
$=\frac{35}{7}$
$=5$
View full question & answer→Question 1222 Marks
If two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are such that $|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=1$ and $\vec{\text{a}}.\vec{\text{b}}=1,$ then find the value of $\big(3\vec{\text{a}}-5\vec{\text{b}}\big).\big(2\vec{\text{a}}+7\vec{\text{b}}\big).$
Answer$\big(3\vec{\text{a}}-5\vec{\text{b}}\big).\big(2\vec{\text{a}}+7\vec{\text{b}}\big)$
$=3\vec{\text{a}}.2\vec{\text{a}}+3\vec{\text{a}}.7\vec{\text{b}}-5\vec{\text{b}}.2\vec{\text{a}}-5\vec{\text{b}}.7\vec{\text{b}}$
$=6\vec{\text{a}}.\vec{\text{a}}+21\vec{\text{a}}.\vec{\text{b}}-10\vec{\text{a}}.\vec{\text{b}}-35\vec{\text{b}}.\vec{\text{b}}$
$=6|\vec{\text{a}}|^2+11\vec{\text{a}}.\vec{\text{b}}-35\big|\vec{\text{b}}\big|^2$
$=6\times2^2+11\times1-35\times1^2$
$=35-35$
$=0$
View full question & answer→Question 1232 Marks
Find the manitude of two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ that are of the same magnitude, are inclined at 60° and whose scalar product is $\frac{1}{2}.$
AnswerGiven that the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $30^{\circ}.$
Also,
$|\vec{\text{a}}|=\big|\vec{\text{b}}\big|;\vec{\text{a}}.\vec{\text{b}}=\frac{1}{2}$
We know that
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow\frac{1}2{}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos60$
$\Rightarrow\frac{1}{2}=|\vec{\text{a}}|^2\big(\frac{1}{2}\big)$
$\Rightarrow|\vec{\text{a}}|^2=1$
$\Rightarrow|\vec{\text{a}}|=1$
$\therefore|\vec{\text{a}}|=\big|\vec{\text{b}\big|}=1$
View full question & answer→Question 1242 Marks
Find a vector in the direction of $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$, which has magnitude of 6 units.
AnswerGiven:
$\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$|\vec{\text{a}}|=\sqrt{2^2+(-1)^2+2^2}$
$=\sqrt{4+1+4}$
$=\sqrt{9}$
$=3$
$\therefore$ Required Vector $=6\times\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=6\times\frac{\big(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big)}3$
$=4\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}$
View full question & answer→Question 1252 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 1262 Marks
Write the value of $\big[2\hat{\text{i }} 3\hat{\text{j }}4\hat{\text{k}}\big].$
AnswerWe have
$\big[2\hat{\text{i }}3\hat{\text{j }}4\hat{\text{k}}\big]$
$=\big(2\hat{\text{i}}\times3\hat{\text{j}}\big).4\hat{\text{k}}$ $\big(\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big)$
$=6\hat{\text{k}}.4\hat{\text{k}}$
$=24$
View full question & answer→Question 1272 Marks
Find the components along the coordinate axis of the position vector of the following point:R(-11, -9)
AnswerHere, R = (-11, -9)
Position vector of $\text{R}=-11\hat{\text{i}}-9\hat{\text{j}}$
Component of R along x-axis $=-11\hat{\text{i}}$
Component of R along x-axis $=-9\hat{\text{j}}$
View full question & answer→Question 1282 Marks
Write two different vectors having same magnitude.
AnswerConsider $\vec{a}=(\hat{i}-2\hat{j}+3\hat{k})\ \text{and}\ \vec{b}=(2\hat{i}+\hat{j}-3\hat{k}).$ It can be observed that $\Big|\vec{a}\Big|=\sqrt{1^2+(-2)^2+3^2}=\sqrt{1+ 4+9}=\sqrt{14}\ \text{and}$$\Big|\vec{b}\Big|=\sqrt{2^2+1^2+(-3)^2}=\sqrt{4+ 1+9}=\sqrt{14}.$
Hence, $\vec{a}\ \text {and}\ \vec{b}$ are two different vectors having the same magnitude. The vectors are different because they have different directions.
View full question & answer→Question 1292 Marks
$\text{Find}\ \lambda\ \text{and}\ \mu\ \text{if}\ (2\hat{i}+6\hat{j}+27\hat{k})\times(\hat{i}+\lambda\hat{j}+\mu\hat{k})=\vec{0}.$
Answer$\text{Given:}\ \ (2\hat{i}+6\hat{j}+27\hat{k})\times(\hat{i}+\lambda\hat{j}+\mu\hat{k})=0$ $\Rightarrow\ \ \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&6&27\\1&\lambda&\mu\end{vmatrix}=\vec{0}$ Expanding along first row, $\hat{i}(6\mu-27\lambda)-\hat{j}(2\mu-27)+\hat{k}(2\lambda-6)$ $=\vec{0}=0\hat{i}+0\hat{j}+0\hat{k}$ Comparing the coefficients of $\hat{i}, \hat{j},\hat{k}$ on both sides, we have$6\mu-27\lambda=0\ \ \ \ \ .....\text{(i)}$
$2\mu-27=0\ \ \ \ \ \ .....\text{(ii)}$
$\text{And}\ \ 2\lambda-6=0\ \ \ \ \ \ .....\text{(iii)}$ $\text{From eq. (ii),}\ \ 2\mu-27=0 \ \Rightarrow\ \ \mu=\frac{27}{2}$ $\text{From eq. (iii),}\ \ 2\lambda-6=0\ \Rightarrow\ \ \lambda=\frac{6}{2}=3$ Putting the values of $\mu\ \text{and}\ \lambda$ in eq. (i), $6\bigg(\frac{27}{2}\bigg)-27(3)=0\ $ $ \Rightarrow\ 3(27)-27(3)=0\ \Rightarrow\ \ 0=0$ $\text{Therefore,}\ \ \mu=\frac{27}{2}\ \text{and}\ \lambda=\frac{6}{2}=3$
View full question & answer→Question 1302 Marks
If $\vec{\text{a}},\vec{\text{b}}$ are non-collinear vectors, then find the value of $\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{i}}\big]\hat{\text{i}}+\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{j}}\big]\hat{\text{j}}+\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{k}}\big]\hat{\text{k}}.$
AnswerFor any vector $\vec{\text{r}},$ we have
$\big(\vec{\text{r}}.\hat{\text{i}}\big)\hat{\text{i}}+\big(\vec{\text{r}}.\hat{\text{j}}\big)\hat{\text{j}}+\big(\vec{\text{r}}.\hat{\text{k}}\big)\hat{\text{k}}=\vec{\text{r}}$
Replacing $\vec{\text{r}}$ by $\vec{\text{a}}\times\vec{\text{b}},$ we have
$\big[(\vec{\text{a}}\times\vec{\text{b}}\big).\hat{\text{i}}\big]+\big[\big(\vec{\text{a}}\times\vec{\text{b}}\big).\hat{\text{j}}+\big[\big(\vec{\text{a}}\times\vec{\text{b}}\big).\hat{\text{k}}\big]\hat{\text{k}}=\vec{\text{a}}\times\vec{\text{b}}$
$\therefore\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{i}}\big]\hat{\text{i}}\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{j}}\big]\hat{\text{j}}+\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{k}}\big]\hat{\text{k}}=\vec{\text{a}}\times\vec{\text{b}}$
View full question & answer→Question 1312 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}$
View full question & answer→Question 1322 Marks
If $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are non-coplanar vectors, prove that the given vectors are non-coplanar:
$2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}},\ \vec{\text{a}}+\vec{\text{b}}-2\vec{\text{c}}$ and $\vec{\text{a}}+\vec{\text{b}}-3\vec{\text{c}}$
AnswerLet if possible the given vectors are coplanar. Then one of the vector is expressible in the terms of the other two.
We have,
$2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}=\text{x}\big(\vec{\text{a}}+\vec{\text{b}}-2\vec{\text{c}}\big)+\text{y}\big(\vec{\text{a}}+\vec{\text{b}}-3\vec{\text{c}}\big)$
$=\vec{\text{a}}(\text{x + y})+\vec{\text{b}}(\text{x + y})+\vec{\text{c}}(-2\text{x}-3\text{y})$
$\Rightarrow\text{x + y}=2,\ \text{x + y}=-1,\ -2\text{x}-3\text{y}=3$
which is not true, as $\text{x + y}=2\neq-1$. Hence, the given vectors are non-coplanar.
View full question & answer→Question 1332 Marks
Find the components along the coordinate axis of the position vector of the following point:Q(-5, 1)
AnswerHere, Q = (-5, 1)
Position vector of $\text{Q}=-5\hat{\text{i}}+\hat{\text{j}}$
Component of Q along x-axis $=-5\hat{\text{i}}$
Component of Q along x-axis $=\hat{\text{j}}$
View full question & answer→Question 1342 Marks
Find $\big|\vec{\text{a}}-\vec{\text{b}}\big|$ if
$|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=5$ and $\vec{\text{a}}.\vec{\text{b}}=8$
AnswerGiven that$|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=5$ and $\vec{\text{a}}.\vec{\text{b}}=8\dots(1)$
We know that
$\big|\vec{\text{a}}-\vec{\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(8)$ [using (1)]
$=4+25-16$
$=13$
$\therefore \big|\vec{\text{a}}-\vec{\text{b}}\big|=\sqrt{13}$
View full question & answer→Question 1352 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 1362 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 1372 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}.$
AnswerGiven: 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 1382 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 1392 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 1402 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 1412 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 1422 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 1432 Marks
Find the angle at which the following vectors are inclined to each of the coordinate axes:
$4\hat{\text{i}}+8\hat{\text{j}}+\hat{\text{k}}$
AnswerLet $\vec{\text{r}}$ be the given vector, and let it make an angle $\alpha,\beta,\gamma$ with OX, OY, OZ respectively. Then, its direction cosines are $\cos\alpha,\cos\beta,\cos\gamma$.So direction ratios of $\vec{\text{r}}=4\hat{\text{i}}+8\hat{\text{j}}+\hat{\text{k}}$ are proportional to 4, 8, 1. Therefore,
Direction cosine of $\vec{\text{r}}$ are $\frac{4}{\sqrt{4^2+8^2+1^2}},\frac{8}{\sqrt{4^2+8^2+1^2}},\frac{1}{\sqrt{4^2+8^2+1^2}}$ or $\frac{4}{9},\frac{8}{9},\frac{1}{9}.$
$\therefore\alpha=\cos^{-1}=\Big(\frac{4}{9}\Big),\beta=\cos^{-1}=\Big(\frac{8}{9}\Big),\gamma=\cos^{-1}=\Big(\frac{1}{9}\Big)$.
View full question & answer→Question 1442 Marks
Find $\vec{\text{a}}.\vec{\text{b}}$ when
$\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=4\hat{\text{i}}-4\hat{\text{j}}+7\hat{\text{k}}$
Answer$\vec{\text{a}}.\vec{\text{b}}$
$=(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}).(4\hat{\text{i}}-4\hat{\text{j}}+7\hat{\text{k}})$
$=(1)(4)+(-2).(-4)+(1)(7)$
$=4+8+7$
$=19$
$\vec{\text{a}}.\vec{\text{b}}=19$
View full question & answer→Question 1452 Marks
Find $\big|\vec{\text{a}}-\vec{\text{b}}\big|$ if
$|\vec{\text{a}}|=3,\big|\vec{\text{b}}\big|=4$ and $\vec{\text{a}}.\vec{\text{b}}=1$
AnswerGiven that$|\vec{\text{a}}|=3,\big|\vec{\text{b}}\big|=4$ and $\vec{\text{a}}.\vec{\text{b}}=1\dots(1)$
We know that
$\big|\vec{\text{a}}-\vec{\text{b}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}.\vec{\text{b}}$
$=3^2+4^2-2(1)$ [using (1)]
$=9+16-2$
$=23$
$\therefore \big|\vec{\text{a}}-\vec{\text{b}}\big|=\sqrt{23}$
View full question & answer→Question 1462 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors of the same magnitude inclined at an angle of $60^\circ$ such that $\vec{\text{a}}.\vec{\text{b}}=8,$ write the value of their magnitude.
AnswerGiven that
$|\vec{\text{a}}|=\big|\vec{\text{b}}\big|$
and $\vec{\text{a}}$ and $\vec{\text{b}}$ are inclined at an angle of $60^\circ$
Also, given that
$\vec{\text{a}}.\vec{\text{b}}=8$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos60^\circ=8$
$\Rightarrow|\vec{\text{a}}||\vec{\text{a}}|\big(\frac{1}{2}\big)=8$
$\Rightarrow|\vec{\text{a}}|^2=16$
$\Rightarrow|\vec{\text{a}}|=4$
View full question & answer→Question 1472 Marks
Find the sum of the vectors $\vec{a}=\hat{i}-2\hat{j}+\hat{k,}\ \ \vec{b}=-2\hat{i}+4\hat{j}+5\hat{k}\ \text{and}\ \vec{c}= \hat{i}-6\hat{j}-7\hat{k}.$
AnswerThe given vectors are $\vec{a}=\hat{i}-2\hat{j}+\hat{k,}\ \ \vec{b}=-2\hat{i}+4\hat{j}+5\hat{k}\ \text{and}\ \vec{c}= \hat{i}-6\hat{j}-7\hat{k}$ $\therefore\vec{a}+\vec{b}+\vec{c}=(1-2+1)\hat{i}+(-2+4-6)\hat{j}+(1+5-7)\hat{k}$$=0\cdot\hat{i}-4\hat{j}-1\cdot\hat{k}$
$=-4\hat{j}-\hat{k}$
View full question & answer→Question 1482 Marks
Find the sum of the following vectors: $\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}},\vec{\text{b}}=2\hat{\text{i}}-3\hat{\text{j}},\vec{\text{c}}=2\hat{\text{i}}-3\hat{\text{k}}$.
AnswerGiven: $\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}},\vec{\text{b}}=2\hat{\text{i}}-3\hat{\text{j}},\vec{\text{c}}=2\hat{\text{i}}-3\hat{\text{k}}$So, Sum of the three vectors$=\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{i}}+3\hat{\text{k}}$
$=5\hat{\text{i}}-5\hat{\text{j}}+3\hat{\text{k}}$
View full question & answer→Question 1492 Marks
$\text{If}\ \vec{a}\cdot\vec{a}=0\ \text{and}\ \vec{a}\cdot\vec{b}=0,$ then what can be concluded about the vector $\vec{b}?$
Answer$\text{Given:}\ \ \vec{a}.\vec{a}=0\Rightarrow\ \ \big|\vec{a}\big|^2=0$ $ \ \Rightarrow\ \ \ \ \big|\vec{a}\big|=0$ $\text{Again}\ \ \vec{a}.\vec{b}=0\ \Rightarrow\ \ \big|\vec{a}\big|.\Big|\vec{b}\Big|\text{cos}\ \theta=0$ $\ \Rightarrow\ \ 0.\Big|\vec{b}\Big|\text{cos}\theta=0\ \Big[\because\big|\vec{a}\big|=0\Big]$ $\Rightarrow\ $ 0 = 0 for all (any vector $\vec{b}.$)Therefore, $\vec{b}$ can be any vector.
View full question & answer→Question 1502 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are vectors of equal magnitude, write the value of $\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big).$
AnswerWE have
$|\vec{\text{a}}|=\big|\vec{\text{b}}\big|\dots(1)$
Now,
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)$
$=|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2$
$=|\vec{\text{a}}|^2-|\vec{\text{a}}|^2$ [using (1)]
$=0$
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