Question 13 Marks
Find a unit vector perpendicular to both the vectors $4\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ and $-2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}.}$
AnswerA vector perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}=\vec{\text{a}}\times\vec{\text{b}}.$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2 \end{vmatrix}$
$\vec{\text{c}}(\text{say})=\begin{bmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\4&-1&3\\-2&1&-2 \end{bmatrix}$
$\vec{\text{c}}=\hat{\text{i}}(2-3)-\hat{\text{j}}(-8+6)+\hat{\text{k}}(4-2)$
$\vec{\text{c}}=-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{c}}$ is a vector perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}.$
$\hat{\text{c}}=\frac{\vec{\text{c}}}{|\vec{\text{c}}|}$
$=\frac{-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}}{\sqrt{(-1)^2+(2)^2+(2)^2}}$
$=\frac{-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}}{\sqrt{1+4+4}}$
$=\frac{1}{3}\big(-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)$
So, unit vector perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}=\frac{1}{3}\big(-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big).$
View full question & answer→Question 23 Marks
Write the number of vectors of unit length perpendicular to both the vectors $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$and $\vec{\text{b}}=\hat{\text{j}}+\hat{\text{k}}.$
AnswerUnit vectors perpendicular to $\vec{\text{a}}$ and $\vec{\text{b}}$ are $\pm\Bigg(\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\times\vec{\text{b}}\big|}\Bigg).$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&1&2\\0&1&1\end{vmatrix}=-\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$
$\therefore$ Unit vectors perpendicular to $\vec{\text{a}}$ and $\vec{\text{b}}$ are $\pm\frac{-\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}}{\sqrt{(-1)^2+(-2)^2+(2)^2}}=\pm\Big(-\frac{1}{3}\hat{\text{i}}-\frac{2}{3}\hat{\text{j}}+\frac{2}{3}\hat{\text{k}}\Big)$
Thus, there are two unit vectors perpendicular to the given vectors.
View full question & answer→Question 33 Marks
If $|\vec{\text{a}}|=\sqrt{26,}\big|\vec{\text{b}}\big|=7$ and $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=35,$ find $\vec{\text{a}}.\vec{\text{b}}.$
Answer$\vec{\text{a}}\times\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta.\hat{\text{n}}$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big||\sin\theta||\hat{\text{n}}|$
$35=\sqrt{26.7}|\sin\theta|.1$
$\sin\theta=\frac{35}{\sqrt{26.5}}$
$\sin\theta=\frac{5}{\sqrt{26}}$
$\cos^2\theta=1-\sin^2\theta$
$=1-\Big(\frac{5}{\sqrt{26}}\Big)^2$
$=\frac{1}{1}-\frac{25}{26}$
$=\frac{26-25}{26}$
$=\frac{1}{26}$
$\cos\theta=\frac{1}{\sqrt{26}}$
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$=\sqrt{26}.7.\frac{1}{\sqrt{26}}$
$\vec{\text{a}}.\vec{\text{b}}=7$
View full question & answer→Question 43 Marks
Find the area of the parallelogram whose diagonals are:
$4\hat{\text{i}}-\hat{\text{j}}-3\hat{\text{k}}$ and $-2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$
AnswerArea of parallalogram $=\frac{1}{2}\big|\vec{\text{d}}_1\times\vec{\text{d}}_2\big|$
Here, $\vec{\text{d}}_1=4\hat{\text{i}}-\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{d}}_2=-2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$
$\vec{\text{d}_1}\times\vec{\text{d}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\4&-1&-3\\-2&1&-2 \end{vmatrix}$
$=\hat{\text{i}}(2+3)-\hat{\text{j}}(-8-6)+\hat{\text{k}}(4-2)$
$=5\hat{\text{i}}+14\hat{\text{j}}+2\hat{\text{k}}$
$\big|\vec{\text{d}}_1\times\vec{\text{d}}_2\big|=\sqrt{(5)^2+(14)^2+(2)^2}$
$=\sqrt{25+196+4}$
$=\sqrt{225}$
$=15$
Area of parallelgram $=\frac{1}{2}\big|\vec{\text{d}}_1\times\vec{\text{d}}_2\big|$
Area of parallelogram $=\frac{15}{2}\text{ sq. unit}$
View full question & answer→Question 53 Marks
If $\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{b}}\times\vec{\text{c}}\neq0,$ then show that $\vec{\text{a}}+\vec{\text{c}}=\text{m}\vec{\text{b}},$ where m is any scalar.
Answer$\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{b}}\times\vec{\text{c}}$$\Rightarrow\vec{\text{a}}\times\vec{\text{b}}=-\vec{\text{c}}\times\vec{\text{b}}$
$\Rightarrow\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{c}}\times\vec{\text{b}}=0$
$\Rightarrow\big(\vec{\text{a}}+\vec{\text{c}}\big)\times\vec{\text{b}}=0$ (using right distributive property)
Thus, $\vec{\text{a}}+\vec{\text{c}}$ is parallel to $\vec{\text{b}}.$
$\Leftrightarrow\vec{\text{a}}+\vec{\text{c}}=\text{m}\vec{\text{b}},$ for som scalar m.
View full question & answer→Question 63 Marks
Define $\vec{\text{a}}\times\vec{\text{b}}$ and prove that $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\big(\vec{\text{a}}.\vec{\text{b}}\big)\tan\theta,$ where $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
AnswerIf $\vec{\text{a}}$ and $\vec{\text{b}}$ are two non-parallel vectors, then the vectors 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\hat{\text{ n}}.$
Here, $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ and $\hat{\text{n}}$ 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 $\hat{\text{n}}$ form a right
$\text{LHS}=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\times\frac{\cos\theta}{\cos\theta}$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\frac{\sin\theta}{\cos\theta}$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\tan\theta$
$=\big(\vec{\text{a}}.\vec{\text{b}}\big)\tan\theta$
$=\text{RHS}$
Hence proved
View full question & answer→Question 73 Marks
Find the area of the parallelogram whose diagonals are:
$2\hat{\text{i}}+\hat{\text{k}}$ and $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
AnswerGiven, $\text{d}_1=2\hat{\text{i}}+\hat{\text{k}}$
$\text{d}_2=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{d}_1}\times\vec{\text{d}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&0&1\\1&1&1 \end{vmatrix}$
$=\hat{\text{i}}(0-1)-\hat{\text{j}}(2-1)+\hat{\text{k}}(2-0)$
$=-\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$\big|\vec{\text{d}}_1\times\vec{\text{d}}_2\big|=\sqrt{(-1)^2+(-1)^2+(2)^2}$
$=\sqrt{1+1+4}$
$=\sqrt{6}$
Area of parallelogram $=\frac{1}{2}\big|\vec{\text{d}}_1\times\vec{\text{d}}_2\big|$
Area of parallelogram $=\frac{1}{2}\sqrt{6}\text{ sq. unit}$
View full question & answer→Question 83 Marks
If either $\vec{\text{a}}=\vec{0}$ or $\vec{\text{b}}=\vec{0},$ then $\vec{\text{a}}\times\vec{\text{b}}=\vec{0}.$ is the converse true? justify your answer with an example.
AnswerIf $\vec{\text{a}}=\vec{0}$ or $\vec{\text{b}}=\vec{0},$ then $|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\text{ n}=\vec{0}.$
$\Rightarrow\vec{\text{a}}\times\vec{\text{b}}=\vec{0}$
But the converse is not true as whenever $\vec{\text{a}}\times\vec{\text{b}}=\vec{0},$ we cannot be sure that either $\vec{\text{a}}=\vec{0}$ or $\vec{\text{b}}=\vec{0}.$
For exampale:
$\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}+3\vec{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+3\vec{\text{k}}$
Here,
$\vec{\text{a}}\neq0$
$\vec{\text{b}}\neq0$
But $\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&2&3\\1&2&3 \end{vmatrix}$
$=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$
$=\vec{0}$
View full question & answer→Question 93 Marks
Find the area of the parallelogram determinrd by the vectors:
$\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}$ and $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}.$
View full question & answer→Question 103 Marks
If $\vec{\text{a}}=3\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}},$ find $\big(\vec{\text{a}}+2\vec{\text{b}}\big)\times\big(\vec{\text{a}}-2\vec{\text{b}}\big).$
AnswerGiven: $\vec{\text{a}}=3\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}$ $\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$$\therefore\vec{\text{a}}+2\vec{\text{b}}=3\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}+2\big(2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}\big)$
$=7\hat{\text{i}}+5\hat{\text{j}}+0\hat{\text{k}}$
$\therefore2\vec{\text{a}}-\vec{\text{b}}\big(3\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}\big)-\big(2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}\big)$
$=4\hat{\text{i}}-5\hat{\text{j}}-5\hat{\text{k}}$
$\big(\vec{\text{a}}+2\vec{\text{b}}\big)\times\big(2\vec{\text{a}}-\vec{\text{b}}\big)=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\7&5&0\\4&-5&-5 \end{vmatrix}$
$=\hat{\text{i}}(-25+0)-\hat{\text{j}}(-35+0)+\hat{\text{k}}(-35-20)$
$=-25\hat{\text{i}}+35\hat{\text{j}}-55\hat{\text{k}}$
View full question & answer→Question 113 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are three unit vectors such that $\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{c}},\vec{\text{b}}\times\vec{\text{c}}=\vec{\text{a}},\vec{\text{c}}\times\vec{\text{a}}=\vec{\text{b}}.$Show that $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ from an orthonormal right handed triad of unit vectors.
AnswerGiven:
$\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{c}}$
$\vec{\text{b}}\times\vec{\text{c}}=\vec{\text{a}}$
$\vec{\text{c}}\times\vec{\text{a}}=\vec{\text{b}}\dots(1)$
Now,
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{c}}|=1$ ( $\because\vec{\text{c}}$ is a unit vector)
$\big|\vec{\text{b}}\times\vec{\text{c}}\big|=|\vec{\text{a}}|=1$ ($\because\vec{\text{a}}$ is a unit vector)
$\big|\vec{\text{c}}\times\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|=1$ ($\because\vec{\text{b}}$ is a unit vector)
$\therefore\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\big|\vec{\text{b}}\times\vec{\text{c}}\big|=\big|\vec{\text{c}}\times\vec{\text{a}}\big|=1\dots(2)$
From (1) and (2), we know
$\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ form an orthonormal right handed triad of unit vectors.
View full question & answer→Question 123 Marks
For any two vectors $\vec{\text{a}}$ and $\vec{\text{b}},$ prove that $\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2=\begin{vmatrix}\vec{\text{a}}.\vec{\text{a}}&\vec{\text{a}}.\vec{\text{b}}\\\vec{\text{b}}.\vec{\text{a}}&\vec{\text{b}} .\vec{\text{b}}\end{vmatrix}.$
Answer$\text{RHS}=\begin{vmatrix}\vec{\text{a}}.\vec{\text{a}}&\vec{\text{a}}.\vec{\text{b}}\\\vec{\text{b}}.\vec{\text{a}}&\vec{\text{b}} .\vec{\text{b}}\end{vmatrix}.$
$=\begin{vmatrix}|\vec{\text{a}}|^2&|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\\|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta&\big|\vec{\text{b}}\big|^2 \end{vmatrix}$
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2-|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2\cos^2\theta$
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2(1-\cos^2\theta)$
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2\sin^2\theta$
$=\big(|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\big)^2$
$=\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2$
$=\text{LHS}$
Hence proved.
View full question & answer→Question 133 Marks
Find a vactor of magnitude $\sqrt{171}$ which is perpendicular to both of the vectors $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$ and $\vec{\text{b}}=3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}.$
AnswerThe Given vectors are $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$ and $\vec{\text{b}}=3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
Unit vectors perpenticular to both $\vec{\text{a}}$ and $\vec{\text{b}}=\pm\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\times\vec{\text{b}}\big|}$
Now,
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&2&-3\\3&-1&2\end{vmatrix}=\hat{\text{i}}-11\hat{\text{j}}-7\hat{\text{k}}$
$\therefore\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\big|\hat{\text{i}}-11\hat{\text{j}}-7\hat{\text{k}}\big|$
$=\sqrt{1^2+(-11)^2+(-7)^2}$
$=\sqrt{1+121+49}$
$=\sqrt{171}$
Unit vectors perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}=\pm\frac{\hat{\text{i}}-11\hat{\text{j}}-7\hat{\text{k}}}{\sqrt{171}}$
Required vectors $=\sqrt{171}\Big(\pm\frac{\hat{\text{i}}-11\hat{\text{j}}-7\hat{\text{k}}}{\sqrt{171}}\Big)=\pm\big(\hat{\text{i}}-11\hat{\text{j}}-7\hat{\text{k}}\big)$
Thus, the vectors of magnitude $\sqrt{171}$ Which are perpendicular to both the given vector are $\pm\big(\hat{\text{i}}-11\hat{\text{j}}-7\hat{\text{k}}\big).$
View full question & answer→Question 143 Marks
What inference can you draw if $\vec{\text{a}}\times\vec{\text{b}}=\vec{0}$ and $\vec{\text{a}}.\vec{\text{b}}=0.$
AnswerGiven:
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\vec{0}$
$\Rightarrow\vec{\text{a}}=0$
$\vec{\text{b}}=0$
$\therefore\vec{\text{a}}||\vec{\text{b}}$
Also,
$\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=0$
$\Rightarrow\vec{\text{a}}=\vec{0}$ or $\vec{\text{b}}=\vec{0}$ or, $\vec{\text{a}}\perp\vec{\text{b}}$
But $\vec{\text{a}}$ cannot be both perpendicular as well as parallel to $\vec{\text{b}}.$
$\therefore|\vec{\text{a}}|=0$
$\big|\vec{\text{b}}\big|=0$
View full question & answer→Question 153 Marks
Find the area of the parallelogram whose diagonals are:
$3\hat{\text{i}}+4\hat{\text{j}}$ and $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
AnswerGiven, $\text{d}_1=3\hat{\text{i}}+4\hat{\text{j}}$
$\text{d}_2=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{d}_1}\times\vec{\text{d}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&4&0\\1&1&1 \end{vmatrix}$
$=\hat{\text{i}}(4-0)-\hat{\text{j}}(3-0)+\hat{\text{k}}(3-4)$
$=4\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}}$
$\big|\vec{\text{d}}_1\times\vec{\text{d}}_2\big|=\sqrt{(4)^2+(-3)^2+(-1)^2}$
$=\sqrt{16+9+1}$
$=\sqrt{26}$
Area of parallelogram $=\frac{1}{2}\big|\vec{\text{d}}_1\times\vec{\text{d}}_2\big|$
Area of parallelogram $=\frac{\sqrt{26}}{2}\text{ sq. unit}$
View full question & answer→Question 163 Marks
Find a vector of magnitude 49, which is perpendicular to both the vectors $2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$ and $3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}.$
AnswerLet, $\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}},\vec{\text{b}}=3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}$
If $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{b}_1\hat{\text{j}}+\text{c}_1\hat{\text{k}}$ and
$\vec{\text{b}}=\text{a}_2\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{c}_2\hat{\text{k}},$ then,
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2 \end{vmatrix}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&3&6\\3&-6&2 \end{vmatrix}$
$=\hat{\text{i}}(6+36)-\hat{\text{j}}(4-18)+\hat{\text{k}}(-12-9)$
$=42\hat{\text{i}}+14\hat{\text{j}}-21\hat{\text{k}}$
$=7\big(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=7\sqrt{(6)^2+(2)^2(-3)^2}$
$=7\sqrt{36+4+9}$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=7\sqrt{49}$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=7\times7$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=49$
View full question & answer→Question 173 Marks
Find the area of the parallelogram determinrd by the vectors:
$2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}$ and $\hat{\text{i}}-\hat{\text{j}}$
AnswerLet:
$\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}+0\hat{\text{k}}$
$\therefore\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&1&3\\1&-1&0 \end{vmatrix}$
$=(0+3)\hat{\text{i}}-(0-3)\hat{\text{j}}+(-2-1)\hat{\text{k}}$
$=3\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}}$
Area of the parallelogram $=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$=\sqrt{3^2+3^2+3^2}$
$=\sqrt{27}$
$=3\sqrt{3}\text{ sq. units}$
View full question & answer→Question 183 Marks
Find the area of the parallelogram determinrd by the vectors:
$3\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and $\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
AnswerLet:$\vec{\text{a}}=3\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$
$\vec{\text{b}}=1\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&1&-2\\1&-3&4 \end{vmatrix}$
$=\hat{\text{i}}(4-6)-\hat{\text{j}}(12+2)+\hat{\text{k}}(-9-1)$
$=-2\hat{\text{i}}-14\hat{\text{j}}-10\hat{\text{k}}$
Area of the parallelogram $=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$=\sqrt{(-2)^2+(-14)^2+(-10)^2}$
$=\sqrt{300}$
$=10\sqrt{3}\text{ sq. units}$
View full question & answer→Question 193 Marks
If $\vec{\text{a}}=4\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{k}},$ then find $\big|2\hat{\text{b}}\times\vec{\text{a}}\big|.$
AnswerGiven:
$\vec{\text{a}}=4\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$
$2\vec{\text{b}}=2\hat{\text{i}}+0\hat{\text{j}}+4\hat{\text{k}}$
$2\vec{\text{b}}\times\vec{\text{a}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&0&-4\\4&3&1 \end{vmatrix}$
$=(0+12)\hat{\text{i}}-(2+16)\hat{\text{j}}+(6-0)\hat{\text{k}}$
$=12\hat{\text{i}}-18\hat{\text{j}}+6\hat{\text{k}}$
$\Rightarrow\big|2\vec{\text{b}}\times\vec{\text{a}}\big|=\sqrt{12^2+(-18^2)+6^2}$
$=\sqrt{504}$
View full question & answer→Question 203 Marks
Find the area of the triangle formed by O, A, B when $\overrightarrow{\text{OA}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\overrightarrow{\text{OB}}=-3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}.$
AnswerGiven: $\overrightarrow{\text{OA}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ $\overrightarrow{\text{OB}}=-3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ $\overrightarrow{\text{OA}}\times\overrightarrow{\text{OB}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&2&3\\-3&-2&1 \end{vmatrix}$ $=8\hat{\text{i}}-10\hat{\text{j}}+4\hat{\text{k}}$ $\big|\overrightarrow{\text{OA}}\times\overrightarrow{\text{OB}}\big|=\sqrt{64+100+16}$ $=\sqrt{180}$ $=6\sqrt{5}$ Area of the triangle $=\frac{1}{2}\big|\overrightarrow{\text{OA}}\times\overrightarrow{\text{OB}}\big|$ $=\frac{1}{2}(6\sqrt{5})$$=3\sqrt{5}\text{ sq. units}$
View full question & answer→Question 213 Marks
Find the angle between two vectors $\vec{\text{a}}$ and $\vec{\text{b},}$ if $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\vec{\text{a}}.\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|=\vec{\text{a}}.\vec{\text{b}}$
$\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\tan\theta=1$
$\Rightarrow\theta=\frac{\pi}{4}$
View full question & answer→Question 223 Marks
Find the area of the pallallelogram determined by the vectors:$2\hat{\text{i}}$ and $3\hat{\text{j}}$
AnswerLet:
$\vec{\text{a}}=2\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$
$\vec{\text{b}}=0\hat{\text{i}}+3\hat{\text{j}}+0\hat{\text{k}}$
$\therefore\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&0&0\\0&3&0 \end{vmatrix}$
$=(0-0)\hat{\text{i}}-(0-0)\hat{\text{j}}+(6 - 0) \hat{\text{k}}$
$=0\hat{\text{i}}+0\hat{\text{j}}+6\hat{\text{k}}$
Area of the parallelogram $=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$=\sqrt{0+0+6 ^2}$
$=6\text{ sq. units}$
View full question & answer→Question 233 Marks
If $|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=5$ and $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=8,$ 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+8^2=2^2\times5^2$ $\big(\therefore\big|\vec{\text{a}}\times\vec{\text{b}}\big|=8,|\vec{\text{a}}|=2$and $\big|\vec{\text{b}}\big|=5\big)$
$\Rightarrow\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+64=100$
$\Rightarrow\big(\vec{\text{a}}.\vec{\text{b}}\big)^2=36$
$\Rightarrow\big(\vec{\text{a}}.\vec{\text{b}}\big)=6$
View full question & answer→Question 243 Marks
Find the area of the parallelogram whose diagonals are:
$2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$ and $3\hat{\text{i}}-6\hat{\text{i}}+2\hat{\text{k}}$
AnswerHere, $\text{d}_1=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$
$\text{d}_2=3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{d}_1}\times\vec{\text{d}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&3&6\\3&-6&2 \end{vmatrix}$
$=\hat{\text{i}}(6+36)-\hat{\text{j}}(4-18)+\hat{\text{k}}(-12-9)$
$=42\hat{\text{i}}+14\hat{\text{j}}-21\hat{\text{k}}$
$=7\big(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)$
$\big|\vec{\text{d}}_1\times\vec{\text{d}}_2\big|=7\sqrt{(6)^2+(2)^2+(-3)^2}$
$=7\sqrt{36+4+9}$
$=7\sqrt{49}$
$=7\times7$
$=49$
Area of parallelogram $=\frac{1}{2}\big|\vec{\text{d}}_1\times\vec{\text{d}}_2\big|$
Area of parallelogram $=\frac{49}{2}\text{ sq.unit}$
View full question & answer→Question 253 Marks
Find the magnitude of $\vec{\text{a}}=\big(3\hat{\text{k}}+4\hat{\text{j}}\big)\times(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}).$
Answer$\vec{\text{a}}=\big(0\hat{\text{i}}+4\hat{\text{j}}+3\hat{\text{k}}\big)\times\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\0&4&3\\1&1&-1 \end{vmatrix}$
$=\hat{\text{i}}(-4-3)-\hat{\text{j}}(0-3)+\hat{\text{k}}(0-4)$
$=-7\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}}$
$\Rightarrow|\vec{\text{a}}|=\sqrt{(-7)^2+3^2+(-4)^2}$
$=\sqrt{74}$
View full question & answer→Question 263 Marks
If $|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=7$ and $\vec{\text{a}}\times\vec{\text{b}}=3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}},$ find the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
AnswerWe know that, if $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}},$ then $\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\dots(1)$ And, $\vec{\text{a}}\times\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|.\sin\theta.\hat{\text{n}}$ $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|.\big|\vec{\text{b}}\big|.|\sin\theta|.|\hat{\text{n}}|$$=|\vec{\text{a}}|\big|\vec{\text{b}}\big||\sin\theta|.1$ [Since, $\hat{\text{n}}$ is a unit vector]
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\dots(2)$ Given that, $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\vec{\text{a}}.\vec{\text{b}}$ $|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta=|\vec{\text{a}}|.\big|\vec{\text{b}}\big|\cos\theta$ $\sin\theta=\cos\theta$ $\theta=\frac{\pi}{4}$ Angle between $\vec{\text{a}}$ and $\vec{\text{b}}=\frac{\pi}{4}$
View full question & answer→Question 273 Marks
Find a vector whose length is 3 and which is perpendicular to the vector $\vec{\text{a}}=3\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}$ and $\vec{\text{b}}=6\hat{\text{i}}+5\hat{\text{j}}-2\hat{\text{k}}.$
Answervector perpendicular to $\vec{\text{a}}$ and $\vec{\text{b}}$
with magnitude $1=\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\times\vec{\text{b}}\big|}$
$=\frac{1}{49}\big(7\big(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)\big)$
$=\frac{1}{7}\big(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)$
vector of magnitude 49, which is perpendicular to $\vec{\text{a}}$ and $\vec{\text{b}}$
$=49\Bigg(\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\times\vec{\text{b}}\big|}\Bigg)$
$=49\big[\frac{1}{7}\big(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)\big]$
$=42\hat{\text{i}}+14\hat{\text{j}}-21\hat{\text{k}}$
The required vector $=42\hat{\text{i}}+14\hat{\text{j}}-21\hat{\text{k}}$
View full question & answer→Question 283 Marks
If $\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2+\big|\vec{\text{a}}.\vec{\text{b}}\big|^2=400$ and $|\vec{\text{a}}|=5,$ then write the value of $\big|\vec{\text{b}}\big|.$
Answer$\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2+\big|\vec{\text{a}}.\vec{\text{b}}\big|^2=400$ $\Rightarrow\Big\{|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\Big\}^2+\Big\{|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\Big\}^2=400$ $\Rightarrow|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2\sin^2\theta+|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2\cos^2\theta=400$ $\Rightarrow|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2=400$ $\Rightarrow25\times\big|\vec{\text{b}}\big|^2=400$ $\Rightarrow\big|\vec{\text{b}}\big|^2=16$$\Rightarrow\big|\vec{\text{b}}\big|^2=4$
View full question & answer→Question 293 Marks
If $\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}},$ and $\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}},$ then find $\vec{\text{a}}\times\vec{\text{b}}.$ verify that $\vec{\text{a}}$ and $\vec{\text{a}}\times\vec{\text{b}}$ are perpendicular to each other.
AnswerGiven:
$\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}$
$\vec{\text{a}}\times\vec{{\text{b}}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-2&3\\2&3&-5 \end{vmatrix}$
$=\hat{\text{i}}+11\hat{\text{j}}+7\hat{\text{k}}$
Now,
$\vec{\text{a}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)=1-22+21$
$=0$
Thus, $\vec{\text{a}}$ is perpendicular to $\vec{\text{a}}\times\vec{\text{b}}.$
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