Question 1013 Marks
Express $\overrightarrow{\text{AB}}$ in terms of unit vectors $\hat{\text{i}}\text{ and }\hat{\text{j}}$, when the point is:A(-6, 3), B(-2, -5)
Find $\Big|\overrightarrow{\text{AB}}\Big|$
AnswerHere, A = (-6, 3) B = (-2, -5) Position vector of $\text{A}=-6\hat{\text{i}}+3\hat{\text{j}}$ Position vector of $\text{B}=-2\hat{\text{i}}-5\hat{\text{j}}$ $\overrightarrow{\text{AB}}$ = Position vector of B - Position vector of A$=\big(-2\hat{\text{i}}-5\hat{\text{j}}\big)-\big(-6\hat{\text{i}}+3\hat{\text{j}}\big)$
$=-2\hat{\text{i}}-5\hat{\text{j}}+6\hat{\text{i}}-3\hat{\text{j}}$
$\overrightarrow{\text{AB}}=4\hat{\text{i}}-8\hat{\text{j}}$
$\Big|\overrightarrow{\text{AB}}\Big|=\sqrt{(4)^2+(-8)^2}$
$=\sqrt{16+64}$
$=\sqrt{80}$
$=\sqrt{16\times5}$ $=4\sqrt5$$\Big|\overrightarrow{\text{AB}}\Big|=4\sqrt5$
$\overrightarrow{\text{AB}}=4\hat{\text{i}}-8\hat{\text{j}}$
View full question & answer→Question 1023 Marks
If $\vec{\text{a}}$ are $\vec{\text{b}}$ are unit vectors, then find the between $\vec{\text{a}}$ and $\vec{\text{b}},$ given that $\big(\sqrt{3}\vec{\text{a}}-\vec{\text{b}}\big)$ is aunit vector.
AnswerLet the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ be $\theta.$
It is given that $|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=\big|\vec{\text{a}}+\vec{\text{b}}\big|=1.$
$\big|\sqrt{3}\vec{\text{a}}-\vec{\text{b}}\big|=1$
$\Rightarrow\big|\sqrt{3}\vec{\text{a}}-\vec{\text{b}}\big|^2=1$
$\Rightarrow\big|\sqrt{3}\vec{\text{a}}\big|^2-2\sqrt{3}\vec{\text{a}}.\vec{\text{b}}+\big|\vec{\text{b}}\big|^2=1$
$\Rightarrow3|\vec{\text{a}}|^2-2\sqrt{3}|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta+\big|\vec{\text{b}}\big|^2=1$
$\Rightarrow3\times1-2\sqrt{3}\times1\times1\times\cos\theta+1=1$
$\Rightarrow2\sqrt{3}\cos\theta=3$
$\Rightarrow\cos\theta=\frac{\sqrt{3}}{2}=\cos\frac{\pi}{6}$
$\Rightarrow\theta=\frac{\pi}{6}$
Thus, the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\frac{\pi}{6}.$
View full question & answer→Question 1033 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 1043 Marks
Dot product of a vector with $\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}},\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}}$ and $2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}$ are 0, 5 and 8 respectively. Find the vector.
AnswerLet $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$ be the required vector.
Given that
$\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big(\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}\big)=0$
$\Rightarrow\text{a+b}-3\text{c}=0\dots(1)$
$\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big(\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}}\big)=5$
$\Rightarrow\text{a}+3\text{b}-2\text{c}=5\dots(2)$
$\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big(2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}\big)=5$
$\Rightarrow2\text{a+b}+4\text{c}=8\dots(3)$
Solving (1), (2) and (3), we get
$\text{a}=1,\text{b}=2,\text{c}=1$
So, $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
View full question & answer→Question 1053 Marks
If $\vec{\text{a}},\vec{\text{b}}$ are two vectors, then write the truth value of the following statement:$\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|\Rightarrow\vec{\text{a}}=\pm\vec{\text{b}}$
AnswerFalse
$\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|\Rightarrow\vec{\text{a}}=\pm\vec{\text{b}}$
Consider an example,
$\vec{\text{a}}=\text{i}+\sqrt3\text{j}$ and $\vec{\text{b}}=\sqrt2\text{i}+\sqrt2\text{j}$
$\big|\vec{\text{a}}\big|=\sqrt{1^2+\big(\sqrt3\big)^2}=2$ and $\big|\vec{\text{b}}\big|=\sqrt{\big(\sqrt2\big)^2+\big(\sqrt2\big)^2}=2$
Thus, $\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|$ but $\vec{\text{a}}\neq\pm\vec{\text{b}}$
View full question & answer→Question 1063 Marks
Express $\overrightarrow{\text{AB}}$ in terms of unit vectors $\hat{\text{i}}\text{ and }\hat{\text{j}}$, when the point is:A(4, -1), B(1, 3)
Find $\Big|\overrightarrow{\text{AB}}\Big|$
AnswerHere, A = (4, -1) B = (1, 3) Position vector of $\text{A}=4\hat{\text{i}}-\hat{\text{j}}$ Position vector of $\text{B}=\hat{\text{i}}+3\hat{\text{j}}$ $\overrightarrow{\text{AB}}$ = Position vector of B - Position vector of A$=\big(\hat{\text{i}}+3\hat{\text{j}}\big)-\big(4\hat{\text{i}}-\hat{\text{j}}\big)$
$=\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{i}}+\hat{\text{j}}$
$\overrightarrow{\text{AB}}=-3\hat{\text{i}}+4\hat{\text{j}}$
$\Big|\overrightarrow{\text{AB}}\Big|=\sqrt{(-3)^2+(4)^2}$
$=\sqrt{9+16}$
$=\sqrt{25}$
$\Big|\overrightarrow{\text{AB}}\Big|=5$
$\overrightarrow{\text{AB}}=-3\hat{\text{i}}+4\hat{\text{j}}$
View full question & answer→Question 1073 Marks
Find $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]$, when
$\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}},\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{j}}+\hat{\text{k}}$
AnswerGiven:
$\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{c}}=\hat{\text{j}}+\hat{\text{k}}$
$\therefore\vec{\text{a}}\times\vec{\text{b}}=\big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big)\times\big(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
$=\hat{\text{k}}+\hat{\text{j}}+4\hat{\text{k}}+2\hat{\text{i}}+6\hat{\text{j}}-3\hat{\text{i}}$
$=-\hat{\text{i}}+7\hat{\text{j}}+5\hat{\text{k}}$
$\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}=\big(-\hat{\text{i}}+7\hat{\text{j}}+5\hat{\text{k}}\big).\big(\hat{\text{j}}+\hat{\text{k}}\big)$
$=7+5=12\ ...(1)$
Now,
$\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}$
$=12$ [Using (1)]
View full question & answer→Question 1083 Marks
Find the projection of $\vec{\text{b}}+\vec{\text{c}}$ on $\vec{\text{a}},$ where $\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$ and $\vec{\text{c}}=2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}.$
Answerprojection of $\big(\vec{\text{b}}+\vec{\text{c}}\big)$ on $\vec{\text{a}}$
$=\frac{\big(\vec{\text{b}}+\vec{\text{c}}\big).\vec{\text{a}}}{|\vec{\text{a}}|}$
$=\frac{\vec{\text{b}}.\vec{\text{a}}+\vec{\text{c}}.\vec{\text{a}}}{\sqrt{(2)^2+(-2)^2+(1)^2}}$
$=\frac{\big(\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}\big)\big(2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)\big(2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}\big)\big(2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)}{\sqrt{4+4+1}}$
$=\frac{(1)(2)+(2)(-2)+(-2)(1)+(2)(2)+(-1)(-2)+(4)(1)}{\sqrt{9}}$
$=\frac{2-4-2+4+2+4}{3}$
$=\frac{12-6}{3}=\frac{6}{3}=2$
projection of $\big(\vec{\text{b}}+\vec{\text{c}}\big)=2$
View full question & answer→Question 1093 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 1103 Marks
If $\vec{\text{a}}$ are $\vec{\text{b}}$ are two unit vectors such that $\vec{\text{a}}+\vec{\text{b}}$ is $\frac{\pi}{6}.$
AnswerLet the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ be $\theta.$
It is given that $|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=\big|\vec{\text{a}}+\vec{\text{b}}\big|=1.$
$\big|\vec{\text{a}}+\vec{\text{b}}\big|=1$
$\Rightarrow\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=1$
$\Rightarrow|\vec{\text{a}}|^2+2|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta+\big|\vec{\text{b}}\big|^2=1$
$\Rightarrow1+2\times1\times1\times\cos\theta+1=1$
$\Rightarrow2\cos\theta=-1$
$\Rightarrow\cos\theta=-\frac{1}{2}=\cos\frac{2\pi}{3}$
$\Rightarrow\theta=\frac{2\pi}{3}$
View full question & answer→Question 1113 Marks
Find a vector in the direction of vector $5\hat{i}-\hat{j}+2\hat{k}$ which has magnitude 8 units.
Answer$\text{Let}\ \vec{a}=5\hat{i}-\hat{j}+2\hat{k.}$
$\therefore\ {|\vec{a}|}=\sqrt{5^2+(-1)^2+2^2}=\sqrt{25+1+4}=\sqrt{30}$
$\therefore\ {\hat{a}}=\frac{\vec{a}}{|\vec{a}|}=\frac{5\hat{i}-\hat{j}+2\hat{k}}{{\sqrt{30}}}$
Hence, the vector in the direction of vector $5\hat{i}-\hat{j}+2\hat{k}$ which has magnitude 8 units is given by,
$8\hat{a}=8\bigg(\frac{5\hat{i}-{\hat{j}+2\hat{k}}}{\sqrt{30}}\bigg)=\frac{40}{\sqrt{30}}\hat{i}-\frac{8}{\sqrt{30}}\hat{j}+\frac{16}{\sqrt{30}}\hat{k}$
$=8\bigg(\frac{5\vec{i}-\vec{j}+2\vec{k}}{\sqrt{30}}\bigg)$
$=\frac{40}{\sqrt{30}}\vec{i}-\frac{8}{\sqrt{30}}\vec{j}+\frac{16}{\sqrt{30}}\vec{k}$
View full question & answer→Question 1123 Marks
If A, B and C have poition vectors (0, 1, 1), (3, 1, 5) and (0, 3, 3), respectively, show that $\triangle\text{ABC}$ is right-angled at C.
AnswerGive that
$\overrightarrow{\text{OA}}=0\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}};\overrightarrow{\text{OB}}=3\hat{\text{i}}+\hat{\text{j}}+5\hat{\text{k}};\overrightarrow{\text{OC}}=0\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}}$
$\overrightarrow{\text{BC}}=\overrightarrow{\text{OC}}-\overrightarrow{\text{OB}}=-3\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$
$\overrightarrow{\text{CA}}=\overrightarrow{\text{OA}}-\overrightarrow{\text{OC}}=0\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$
Now,
$\overrightarrow{\text{BC}}.\overrightarrow{\text{CA}}=0-4+4=0$
So, $\overrightarrow{\text{BC}}$ is perpendicular to $\overrightarrow{\text{CA}}$.
So, $\triangle\text{ABC}$ is right-angled at C.
View full question & answer→Question 1133 Marks
Find $|\vec{\text{a}|}$ and $\big|\vec{\text{b}}\big|$ if
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=12$ and $|\vec{\text{a}}|=2\big|\vec{\text{b}}\big|$
AnswerHere, $\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=12$
$|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2=12$
$\big(2\big|\vec{\text{b}}\big|\big)^2-\big|\vec{\text{b}}\big|^2=12$ $\big[\text{Using|}\vec{\text{a}}|=2\big|\vec{\text{b}}\big|\big]$
$4\big|\vec{\text{b}}\big|^2-\big|\vec{\text{b}}\big|^2=12$
$3\big|\vec{\text{b}}\big|^2=12$
$\big|\vec{\text{b}}\big|^2=\frac{12}{3}$
$\big|\vec{\text{b}}\big|^2=4$
$\big|\vec{\text{b}}\big|=2$
$\big|\vec{\text{a}}\big|=2\big|\vec{\text{b}}\big|=2(2)$
$\big|\vec{\text{a}}\big|=4$
$\big|\vec{\text{b}}\big|=2$
View full question & answer→Question 1143 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 1153 Marks
Find the value of $\lambda$ so that the following vectors are coplanar:
$\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}},\vec{\text{b}}=3\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}},\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
AnswerGiven:
$\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{b}}=3\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
We know that vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar iff $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0.$
It is given that $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar.
$\therefore\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0$
$\Rightarrow\begin{vmatrix}1&2&-3\\3&\lambda&1\\1&2&2 \end{vmatrix}=0$
$\Rightarrow1(2\lambda-2)-2(6-1)-3(6-\lambda)=0$
$\Rightarrow5\lambda-30=0$
$\Rightarrow\lambda=6$
View full question & answer→Question 1163 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 1173 Marks
If $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}.$ find $\lambda$ such that $\vec{\text{a}}$ is perpendicular to $\lambda\vec{\text{b}}+\vec{\text{c}}.$
AnswerThe given vectors are $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}.$
Now,
$\lambda\vec{\text{b}}+\vec{\text{c}}=\lambda(\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})+(\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})\\=(\lambda+1)\hat{\text{i}}+(\lambda+3)\hat{\text{j}}-(2\lambda+1)\hat{\text{k}}$
It is given
$\vec{\text{a}}\perp\big(\lambda\vec{\text{b}}+\vec{\text{c}}\big)$
$\Rightarrow\vec{\text{a}}.\big(\lambda\vec{\text{b}}+\vec{\text{c}}\big)=0$
$\Rightarrow\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big).\Big[(\lambda+1)\hat{\text{i}}+(\lambda+3)\hat{\text{j}}-(2\lambda+1)\hat{\text{k}}\Big]=0$
$\Rightarrow2(\lambda+1)-(\lambda+3)-(2\lambda+1)=0$
$\Rightarrow2\lambda+2-\lambda-3-2\lambda-1=0$
Thus, the value of $\lambda$ is -2
View full question & answer→Question 1183 Marks
If the vertices A, B, C of a triangle ABC are (1, 2, 3), (-1, 0, 0), (0, 1, 2), respectively, then find $\angle\text{ABC}.[\angle\text{ABC is the angle between the vectors}\ \overrightarrow{\text{BA}}\ \text{and} \ \overrightarrow{\text{BC}}].$
AnswerVertices A, B, C of a triangle are A(1, 2, 3), B(-1, 0, 0) and C(0, 1, 2) respectively. $\therefore\ \ \text{Position vector of point A = }\overrightarrow{\text{OA}} $ $=(1, 2, 3)=\hat{i}+2\hat{j}+3\hat{k}$$\text{Position vector of point B}=\overrightarrow{\text{OB}}$ $=(-1, 0, 0)=-\hat{i}+0\hat{j}+0\hat{k}$
$\text{Position vector of point C}=\overrightarrow{\text{OC}}$ $=(0, 1, 2)=0\hat{i}+1\hat{j}+2\hat{k}$
Now $\ \overrightarrow{\text{BA}}$ = Position vector of point A - Position vector of point B$=\hat{i}+2\hat{j}+3\hat{k}-\big(-\hat{i}+0\hat{j}+0\hat{k}\big)$
$=\hat{i}+2\hat{j}+3\hat{k}+\hat{i}-0\hat{j}-0\hat{k}$ $=2\hat{i}+2\hat{j}+3\hat{k}\ \ \ \ \ \ \ \ \ ......\text{(i)}$
And $ \ \overrightarrow{\text{BC}}$ = Position vector of point C - Position vector of point B$=0\hat{i}+\hat{j}+2\hat{k}-\big(-\hat{i}+0\hat{j}+0\hat{k}\big)$
$=0\hat{i}+\hat{j}+2\hat{k}+\hat{i}-0\hat{j}-0\hat{k}$ $=\hat{i}+\hat{j}+2\hat{k}\ \ \ \ \ \ \ \ \ ......\text{(ii)}$
Let $\theta$ be the angle between the vectors $\overrightarrow{\text{BA}}\ \text{and}\ \overrightarrow{\text{BC}}.$$\therefore\ \ \text{cos}\theta=\frac{\overrightarrow{\text{BA}}.\overrightarrow{\text{BC}}}{\Big|\overrightarrow{\text{BA}}\Big|.\Big|\overrightarrow{\text{BC}}\Big|}=\frac{2(1)+2(1)+3(2)}{\sqrt{4+4+9}\sqrt{1+1+4}}\ $ [Using eq. (i) and (ii)]
$\Rightarrow\ \text{cos}\theta=\frac{10}{\sqrt{17}\sqrt{6}}=\frac{10}{\sqrt{102}}$ $\Rightarrow\ \theta=\cos^{-1}\Big(\frac{10}{\sqrt{102}}\Big)$
View full question & answer→Question 1193 Marks
Find the angle between the vectors $2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $3\hat{\text{i}}+4\hat{\text{j}}-\hat{\text{k}}.$
AnswerLet $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{j}}-\hat{\text{k}}$
We know that, angle between two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ is given by
$\cos\theta=\frac{\vec{\text{a}}\cdot\vec{\text{b}}}{|\vec{\text{a}}||\vec{\text{b}}|}$
$=\frac{(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})(3\hat{\text{i}}+4\hat{\text{j}}-\hat{\text{k}})}{\sqrt{4+1+1}\sqrt{9+16+1}}$
$=\frac{6-4-1}{\sqrt{6}\sqrt{26}}=\frac{1}{2\sqrt{39}}$
$\therefore\theta=\cos^{-1}\Big(\frac{1}{2\sqrt{39}}\Big)$
View full question & answer→Question 1203 Marks
The two adjacent sides of a parallelogram are $2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}\ \text{and}\ \hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}.$ Find the unit vector parallel to its diagonal. Also, find its area.
AnswerAdjacent sides of a parallelogram are given as: $\vec{\text{a}}=2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}\ \text{and}\ \vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
Then, the diagonal of a parallelogram is given by $\vec{\text{a}}+\vec{\text{b}}.$
$\vec{\text{a}}+\vec{\text{b}}=(2+1)\hat{\text{i}}+(-4-2)\hat{\text{j}}+(5-3)\hat{\text{k}}$ $=3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}$
Thus, the unit vector parallel to the diagonal is
$\frac{\vec{\text{a}}+\vec{\text{b}}}{\Big|\vec{\text{a}}+\vec{\text{b}}\Big|}=\frac{3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}}{\sqrt{3^2+(-6)^2+2^2}}=\frac{3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}}{\sqrt{9+36+4}}$ $=\frac{3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}}{7}=\frac{3}{7}\hat{\text{i}}-\frac{6}{7}\hat{\text{j}}+\frac{2}{7}\hat{\text{k}}.$
$\therefore$ Area of parallelogram ABCD $=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&-4&5\\1&-2&-3\end{vmatrix}$
$=\hat{\text{i}}(12+10)-\hat{\text{j}}(-6-5)+\hat{\text{k}}(-4+4)$
$=22\hat{\text{i}}+11\hat{\text{j}}$
$=11\big(2\hat{\text{i}}+\hat{\text{j}}\big)$
$\therefore\big|\vec{\text{a}}\times\vec{\text{b}}\big|=11\sqrt{2^2+1^2}=11\sqrt{5}$
Hence, the area of the parallelogram is $11\sqrt{5}$ square units.
View full question & answer→Question 1213 Marks
Write the projection of $\vec{\text{b}}+\vec{\text{c}}$ on $\vec{\text{a}}$ when $\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$ and $\vec{\text{c}}=2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}.$
AnswerGiven that
$\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$
$\vec{\text{c}}=2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{b}}+\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}+2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}$
$=3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
Projection of $\vec{\text{b}}+\vec{\text{c}}$ on $\vec{\text{a}}$ is
$\frac{\big(\vec{\text{b}}+\vec{\text{c}}\big).\vec{\text{a}}}{|\vec{\text{a}}|}$
$=\frac{\big(3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big).\big(2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)}{2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}}$
$=\frac{6-2+2}{\sqrt{4+4+1}}$
$=\frac{6}{3}$
$=2$
View full question & answer→Question 1223 Marks
Find the unit vector in the direction of the resultant of the vectors $\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}},\ 2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and $\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$.
AnswerGiven: $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}},\ 2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and $\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$ are the position vectors. Then, Resultant of vectors $=\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$$=\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}+2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}+\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$
$=4\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
So, $\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=\sqrt{4^2+2^2+1^2}$$=\sqrt{16+4+1}$
$=\sqrt{21}$
$\therefore$ Unit vector in the direction of the resultant vector $=\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}{\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|}$ $=\frac{1}{\sqrt{21}}\big(4\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\big)$
View full question & answer→Question 1233 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are three non-coplanar vectors, such that $\vec{\text{d}}.\vec{\text{a}}=\vec{\text{d}}.\vec{\text{b}}=\vec{\text{d}}.\vec{\text{c}}=0,$ then show that $\vec{\text{d}}$ is the null vector.
AnswerGiven that
$\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are three non-coplanar vectors such that
$\vec{\text{d}}.\vec{\text{a}}=\vec{\text{d}}.\vec{\text{b}}=\vec{\text{d}}.\vec{\text{c}}=0$
Given that
$\vec{\text{d}}.\vec{\text{a}}=0$
⇒ $\vec{\text{d}}$ perpendicular to $\vec{\text{a}}$
or $\vec{\text{d}}=0\dots(1)$
$\vec{\text{d}}.\vec{\text{b}}=0$
⇒ $\vec{\text{d}}$ is perpendicular to $\vec{\text{b}}$ or $\vec{\text{d}}=0\dots(2)$
$\vec{\text{d}}.\vec{\text{c}}=0$
⇒ $\vec{\text{d}}$ is perpendicular to $\vec{\text{c}}$ or $\vec{\text{d}}=0\dots(3)$
From (1), (2), (3), we get
$\vec{\text{d}}$ is perpendicular to $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ or $\vec{\text{d}}=0,$ but $\vec{\text{d}}$ can not be perpendicular to $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}},$ because $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are three non-coplanar vectors, so
$\vec{\text{d}}=0,$
View full question & answer→Question 1243 Marks
Find $|\vec{\text{a}|}$ and $\big|\vec{\text{b}}\big|$ if
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=3$ and $|\vec{\text{a}}|=2\big|\vec{\text{b}}\big|$
AnswerHere, $\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=3$ $|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2=3$ $\big(2\big|\vec{\text{b}}\big|\big)^2-\big|\vec{\text{b}}\big|^2=3$ $\big[\text{using} |\vec{\text{a}}|=2\big|\vec{\text{b}}\big|\big]$ $4\big|\vec{\text{b}}\big|^2-\big|\vec{\text{b}}\big|^2=3$$3\big|\vec{\text{b}}\big|^2=3$
$\big|\vec{\text{b}}\big|^2=\frac{3}{3}$
$\big|\vec{\text{b}}\big|^2=1$
$\big|\vec{\text{b}}\big|=1$
$|\vec{\text{a}}|=2\big|\vec{\text{b}}\big|$ $=2(1)$ $|\vec{\text{a}}|=2$ $\big|\vec{\text{b}}\big|=1$
View full question & answer→Question 1253 Marks
Find the angle between the vectors $\hat{i}-2\hat{j}+3\hat{k}\ \text{and}\ 3\hat{i}-2\hat{j}+\hat{k}.$
Answer$\text{Given:}\ \ \ \text{Let}\ \ \ \ \vec{a}=\hat{i}-2\hat{j}+3\hat{k}\ \text{and}\ \vec{b}=3\hat{i}-2\hat{j}+\hat{k}$ $\Rightarrow\ \ \big|\vec{a}\big|=\sqrt{1+4+9}=\sqrt{14}\ \text{and}\ \Big|\vec{b}\Big|=\sqrt{9+4+1}=\sqrt{14}$ $\Big[\because\ \text{x}\hat{i}+\text{y}\hat{j}+\text{z}\hat{k}=\sqrt{\text{x}^2+\text{y}^2+\text{z}^2}\ \Big]$ $\text{Also}\ \ \ \ \vec{a}.\vec{b}$= Product of coefficients of $\hat{i}$ + Product of coefficients of $\hat{j}$ + Product of coefficients $\hat{k}$
= 1(3) + (-2)(-2) + 3(1) = 3 + 4 + 3 = 10
Let $\theta$ be the angle between the vector $\vec{a}\ \text{and}\ \vec{b}.$ We know that $\text{cos}\ \theta=\frac{\vec{a}.\vec{b}}{\big|\vec{a}\big|.\big| \vec{b}\big|}$$\Rightarrow\ \ \text{cos}\ \theta=\frac{10}{\sqrt{14}\ .\sqrt{14}}=\frac{10}{14}=\frac{5}{7}$ $ \Rightarrow\ \text{cos}\ \theta=\frac{5}{7}$
$\Rightarrow\ \ \theta=\cos^{-1}\frac{5}{7}$
View full question & answer→Question 1263 Marks
A vector makes an angle of $\frac{\pi}4$ with each of x-axis and y-axis. Find the angle made by it with the z-axis.
AnswerLet the vector $\overrightarrow{\text{OP}}$ makes an angle $\alpha=45^{\circ}$ and $\beta=45^{\circ}$ with OX, OY respectively. Suppose $\overrightarrow{\text{OP}}$ is inclined at angle $\gamma$ to OZ.
Let l, m, n be the direction cosines of $\overrightarrow{\text{OP}}$. Then,
$\text{l}=\cos\frac{\pi}4=\frac{1}{\sqrt2}$
$\text{m}=\cos\frac{\pi}4=\frac{1}{\sqrt2}$
$\text{n}=\cos\gamma$
Now, we have,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\ \frac{1}2+\frac{1}2+\text{n}^2=1$
$\Rightarrow\ \text{n}^2=0$
$\Rightarrow\ \text{n}=0$
$\Rightarrow\ \cos\gamma=\cos\frac{\pi}2$
$\Rightarrow\ \gamma= \frac{\pi}2$
Hence, the angle made by it with the z-axis is $\frac{\pi}2$.
View full question & answer→Question 1273 Marks
Using vectors, find the value of $\lambda$ such that the points ($\lambda$, -10, 3), (1, -1, 3) and (3, 5, 3) are collinear.
AnswerPoints ($\lambda$, -10, 3), (1, -1, 3) and (3, 5, 3) are collinear.$\therefore$ ($\lambda$, -10, 3) = x(1, -1, 3) + y(3, 5, 3) for some scalars x and y.
$\Rightarrow\lambda$ = x + 3y, -10 = -x + 5y and 3 = 3x +3y
Solving -10 = -x + 5y and 3 = 3x + 3y for x and y we get,
$\text{x}=\frac{5}2$ and $\text{y}=-\frac{3}2$
Now,
$\lambda=\text{x}+3\text{y}$
$\Rightarrow\lambda=\frac{5}2+3\Big(-\frac{3}2\Big)=-2$
View full question & answer→Question 1283 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 1293 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 1303 Marks
$\text{If}\ \vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}},$ find a unit vector parallel to the vector $2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}.$
Answer$\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ $2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}=2\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)-\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)+3\big(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)$ $=2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}-2\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}+3\hat{\text{i}}-6\hat{\text{j}}+3\hat{\text{k}}$ $=3\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$ $\Big|2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}\Big|=\sqrt{3^{2}+(-3)^{2}+2^{2}}$ $=\sqrt{9+9+4}=\sqrt{22}$Hence, the unit vector along $2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}\ \text{is}$
$\frac{2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}}{\Big|2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}\Big|}=\frac{3\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}}{\sqrt{22}}$ $\frac{3}{\sqrt{22}}\hat{\text{i}}-\frac{3}{\sqrt{22}}\hat{\text{j}}+\frac{2}{\sqrt{22}}\hat{\text{k}}.$
View full question & answer→Question 1313 Marks
If the position vectors of the points A(3, 4), B(5, -6) and C(4, -1) are $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ respectively, compute $\vec{\text{a}}+2\vec{\text{b}}-3\vec{\text{c}}$.
AnswerHere, A(3, 4), B(5, -6), C(4, -1) $\vec{\text{a}}=3\hat{\text{i}}+4\hat{\text{j}}$ $\vec{\text{b}}=5\hat{\text{i}}-6\hat{\text{j}}$ $\vec{\text{c}}=4\hat{\text{i}}-\hat{\text{j}}$ Now,$\vec{\text{a}}+2\vec{\text{b}}-3\vec{\text{c}}=\big(3\hat{\text{i}}+4\hat{\text{j}}\big)+2\big(5\hat{\text{i}}-6\hat{\text{j}}\big)-3\big(4\hat{\text{i}}-\hat{\text{j}}\big)$
$=3\hat{\text{i}}+4\hat{\text{j}}+10\hat{\text{i}}-12\hat{\text{j}}-12\hat{\text{i}}+3\hat{\text{j}}$ $=\hat{\text{i}}-5\hat{\text{j}}$ $\therefore\ \vec{\text{a}}+2\vec{\text{b}}-3\vec{\text{c}}=\hat{\text{i}}-5\hat{\text{j}}$
View full question & answer→Question 1323 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$$\Rightarrow\big|\vec{\text{b}}\big|^2=4$
View full question & answer→Question 1333 Marks
Using vector method, prove that the point is collinear:
A(2, -1, 3), B(4, 3, 1) and C(3, 1, 2)
AnswerGiven the points A(2, -1, 3), B(4, 3, 1) and C(3, 1, 2). Then, $\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A$=4\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}-2\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}$
$=2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}$
$=-2\big(-\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)$$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}-4\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}}$
$=-\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
$\therefore\ \overrightarrow{\text{AB}}=-2\overrightarrow{\text{BC}}$
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are parallel vectors. But B is a point common to them. Hence, the given points A, B, and C are collinear.
View full question & answer→Question 1343 Marks
Let $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}},\vec{\text{d}}$ be the position vectors of the four distinct points A, B, C, D. If $\vec{\text{b}}-\vec{\text{a}}=\vec{\text{c}}-\vec{\text{d}}$, then show that ABCD is a parallelogram.
AnswerHere it is given that $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}\text{ and }\vec{\text{d}}$ be the position vectors of the four distinct points A, B, C, D such that,
$\vec{\text{b}}-\vec{\text{a}}=\vec{\text{c}}-\vec{\text{d}}$
Given that,
$\vec{\text{b}}-\vec{\text{a}}=\vec{\text{c}}-\vec{\text{d}}$
$\overrightarrow{\text{AB}}=\overrightarrow{\text{DC}}$
So, AB is parallel and equal to DC (in magnitude).
Hence,
ABCD is a parallelogram.
View full question & answer→Question 1353 Marks
Find the value of $\lambda$ so that the following vectors are coplanar:
$\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}++\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}},\vec{\text{c}}=\lambda\hat{\text{i}}+\lambda\hat{\text{j}}+5\hat{\text{k}}$
AnswerGiven:
$\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{c}}=\lambda\hat{\text{i}}+\lambda\hat{\text{j}}+5\hat{\text{k}}$
We know that vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar iff $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0.$
It is given that $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar.
$\therefore\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0$
$\Rightarrow\begin{vmatrix}2&-1&1\\1&2&-3\\\lambda&\lambda&5 \end{vmatrix}=0$
$\Rightarrow2(10+3\lambda)+1(5+3\lambda)+1(\lambda-2\lambda)=0$
$\Rightarrow 8\lambda +25=0$
$=-\frac{25}{8}$
View full question & answer→Question 1363 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 $\vec{\text{a}}\times\vec{\text{b}}.$
View full question & answer→Question 1373 Marks
Find the unit vector in the direction of sum of vectors $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{j}}+\hat{\text{k}}.$
AnswerLet $\vec{\text{c}}$ denote the sum of vectors $\vec{\text{a}}$ and $\vec{\text{b}}.$
Thus $\vec{\text{c}}=\vec{\text{a}}+\vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}+2\hat{\text{j}}+\hat{\text{k}}$
$=2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
Now, we know that, unit vector in the direction of a vector $\vec{\text{a}}$ is given as $\frac{\vec{\text{a}}}{|\vec{\text{a}}|}.$
$\therefore$ unit vector in the direction of $\vec{\text{c}}=\frac{\vec{\text{c}}}{|\vec{\text{c}}|}=\frac{2\hat{\text{i}}+\hat{\text{j}}+\hat{2\text{k}}}{\sqrt{2^2+1^1+2^2}}$
$=\frac{2\hat{\text{i}}+\hat{\text{j}}+\hat{2\text{k}}}{\sqrt{9}}$
Thus, $\vec{\text{c}}=\frac{2\hat{\text{i}}+\hat{\text{j}}+\hat{2\text{k}}}{3}$
View full question & answer→Question 1383 Marks
Using vector method, prove that the point is collinear:
A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1)
AnswerGiven the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1). Then, $\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A$=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}-\hat{\text{i}}-2\hat{\text{j}}-7\hat{\text{k}}$
$=\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}$
$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=3\hat{\text{i}}+10\hat{\text{j}}-\hat{\text{k}}-2\hat{\text{i}}-6\hat{\text{j}}-3\hat{\text{k}}$
$=\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}$
$\therefore\ \overrightarrow{\text{AB}}=\overrightarrow{\text{BC}}$
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are parallel vectors. But B is a point common to them. Hence, the given points A, B, and C are collinear.
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