Question 1512 Marks
If $\vec{\text{a}}=3\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}},\ \vec{\text{b}}=-2\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$, find $|3\vec{\text{a}}-2\vec{\text{b}}+4\vec{\text{c}}|$.
AnswerGiven $\vec{\text{a}}=3\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}},\ \vec{\text{b}}=-2\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}},\ \vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
Now, $3\vec{\text{a}}-2\vec{\text{b}}+4\vec{\text{c}}=3\big(3\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}}\big)-2\big(-2\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}\big)+4\big(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\big)$
$=9\hat{\text{i}}-3\hat{\text{j}}-12\hat{\text{k}}+4\hat{\text{i}}-8\hat{\text{j}}+6\hat{\text{k}}+4\hat{\text{i}}+8\hat{\text{j}}-4\hat{\text{k}}$
$=17\hat{\text{i}}-3\hat{\text{j}}-10\hat{\text{k}}$
$\therefore\ |3\vec{\text{a}}-2\vec{\text{b}}+4\vec{\text{c}}|=\sqrt{17^2+(-3)^2+(-10)^2}$
$=\sqrt{289+9+100}$
$=\sqrt{398}$
View full question & answer→Question 1522 Marks
If $\vec{\text{a}}=\text{x}\hat{\text{i}}+2\hat{\text{j}}-\text{z}\hat{\text{k}}$ and $\vec{\text{b}}=3\hat{\text{i}}-\text{y}\hat{\text{j}}+\hat{\text{k}}$ are two equal vectors, then write the value of x + y + z.
AnswerGiven: $\vec{\text{a}}=\text{x}\hat{\text{i}}+2\hat{\text{j}}-\text{z}\hat{\text{k}}$ and $\vec{\text{b}}=3\hat{\text{i}}-\text{y}\hat{\text{j}}+\hat{\text{k}}$ Since the two vectors are equal. We have,$\vec{\text{a}}=\text{x}\hat{\text{i}}+2\hat{\text{j}}-\text{z}\hat{\text{k}}=\vec{\text{b}}=3\hat{\text{i}}-\text{y}\hat{\text{j}}+\hat{\text{k}}$
$\Rightarrow\ \text{x}=3,\ \text{y}=-2,\ \text{z}=-1$
$\therefore\ \text{x + y + z}=3-2-1=0$
View full question & answer→Question 1532 Marks
If $\vec{\text{a}},\vec{\text{b}}$ are two vectors, then write the truth value of the following statement:$\vec{\text{a}}=-\vec{\text{b}}\Rightarrow\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|$
AnswerTrue
$\vec{\text{a}}=-\vec{\text{b}}$
Take modulus both sides
$\big|\vec{\text{a}}\big|=\big|-\vec{\text{b}}\big|$
$\Rightarrow\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|$ $\Big[\therefore\ \big|-\vec{\text{b}}\big|=\big|\vec{\text{b}}\big|\Big]$
View full question & answer→Question 1542 Marks
$\text{Find}\ \big|\vec{x}\big|,$ if for a unit unit vector $\vec{a},\ (\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a})=12.$
Answer$\text{Given:}\ \vec{a}\ \text{is a unit vector}\ \Rightarrow\ \big|\vec{a}\big|=1\ \ .......(\text{i})$
$\big(\vec{x}-\vec{a}\big)\big(\vec{a}+\vec{a}\big)=12\ $ $\Rightarrow\ \vec{x}.\vec{x}+\vec{x}.\vec{a}-\vec{a}.\vec{x}-\vec{a}.\vec{a}=12$
$\Rightarrow\ \ \big|\vec{x}\big|^2+\vec{x}.\vec{a}-\vec{a}.\vec{x}-\big|\vec{a}\big|^2=12\ $ $\Rightarrow\ \big|\vec{x}\big|^2+\vec{a}.\vec{x}-\vec{a}.\vec{x}-\big|\vec{a}\big|^2=12$
$ \Rightarrow\ \ \big|\vec{x}\big|^2-\big|\vec{a}\big|^2=12$
$\text{Putting}\ \big|\vec{a}\big|=1\ \text{from eq. (i),}\ \ \ \ \big|\vec{x}\big|^2-1=12$
$\Rightarrow\ \big|\vec{x}\big|^2=13\ \ \Rightarrow\ \ \big|\vec{x}\big|=13$
View full question & answer→Question 1552 Marks
If $\vec{\text{a}}\text{ and }\vec{\text{b}}$ denote the position vectors of points A and B respectively and C is a point on AB such that 3AC = 2AB, then write the position vector of C.
AnswerGiven: $\vec{\text{a}}\text{ and }\vec{\text{b}}$ denote the position vectors of points A and B respectively and C is a point on AB such that 3AC = 2AB.Let $\vec{\text{c}}$ is the position vector of C.
Now, $\overrightarrow{\text{AB}}=\vec{\text{b}}-\vec{\text{a}}$ $\overrightarrow{\text{AC}}=\vec{\text{c}}-\vec{\text{a}}$ Consider, $3\text{AC}=2\text{AB}$ $\Rightarrow\ 3\big(\vec{\text{c}}-\vec{\text{a}}\big)=2\big(\vec{\text{b}}-\vec{\text{a}}\big)$ $\Rightarrow\ 3\vec{\text{c}}-3\vec{\text{a}}=2\vec{\text{b}}-2\vec{\text{a}}$ $\Rightarrow\ 3\vec{\text{c}}=2\vec{\text{b}}+\vec{\text{a}}$ $\Rightarrow\ \vec{\text{c}}=\frac{1}3\big(2\vec{\text{b}}+\vec{\text{a}}\big)$ $\Rightarrow\ \vec{\text{c}}=\frac{1}3\big(\vec{\text{a}}+2\vec{\text{b}}\big)$ Hence, the position vector of C is $\frac{1}3\big(\vec{\text{a}}+2\vec{\text{b}}\big)$
View full question & answer→Question 1562 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 1572 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 1582 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 1592 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 1602 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 1612 Marks
A unit vector $\vec{\text{r}}$ makes angles $\frac{\pi}3\text{ and }\frac{\pi}2$ with $\hat{\text{j}}\text{ and } \hat{\text{k}}$ respectively and an acute angle $\theta$ with $\hat{\text{i}}$. Find $\theta$.
AnswerA unit vector makes an angle $\frac{\pi}3\text{ and }\frac{\pi}2$ with $\hat{\text{j}}\text{ and } \hat{\text{k}}$Let l, m, n be its direction cosines
$\therefore\ \text{l}=\cos\theta,\ \text{m}=\cos\big(\frac{\pi}3\big)=\frac{1}2,\ \text{n}=\cos\big(\frac{\pi}2\big)=0$
Now,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\ \text{l}^2+\frac{1}4+0=1$
$\Rightarrow\ \text{l}^2=1-\frac{1}4=\frac{3}4$
$\Rightarrow\ \text{l}=\pm\frac{\sqrt3}2$
$\therefore\ \vec{\text{r}}$ makes an acute angle 30º, 150º with $\hat{\text{i}}$
Since, angle $\theta$ is acute.
$\therefore\ \theta=30^{\circ}$
View full question & answer→Question 1622 Marks
Write the position vector of a point dividing the line segment joining points having position vectors $\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and $2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ externally in the ratio 2 : 3.
AnswerLet A and B be the points with position vectors $\vec{\text{a}} = \hat{\text{i}} + \hat{\text{j}} - 2\hat{\text{k}},\vec{\text{b}} = 2\hat{\text{i}} - \hat{\text{j}} + 3\hat{\text{k}} $ respectively.
Let C divide AB externally in the ratio 2 : 3 such that AC : CB = 2 : 3
$\therefore $ Postion vector of C $ = \frac{2\big(2\hat{\text{i}} - \hat{\text{j}} + 3\hat{\text{k}}\big) - 3 \big(\hat{\text{i}} + \hat{\text{j}} - 2\hat{\text{k}}\big)}{2 - 3}$
$ = \frac{4\hat{\text{i}} - 2\hat{\text{j}} + 6\hat{\text{k}} - 3\hat{\text{i}} - 3\hat{\text{j}} + 6\hat{\text{k}}}{-1}$
$= \frac{\hat{\text{i}} - 5\hat{\text{j}} + 12\hat{\text{k}}}{-1}$
$= -\hat{\text{i}} + 5\hat{\text{j}} - 12\hat{\text{k}}$
View full question & answer→Question 1632 Marks
Write a unit vector in the direction of $\overrightarrow{\text{PQ}}$, where P and Q are the points (1, 3, 0) and (4, 5, 6) respectively.
AnswerP(1, 3, 0) and Q(4, 5, 6) are the given points.
$\therefore\ \overrightarrow{\text{PQ}}=\big(4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}\big)-\big(\hat{\text{i}}+3\hat{\text{j}}+0\hat{\text{k}}\big)$
$=3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$
$\Rightarrow\ \Big|\overrightarrow{\text{PQ}}\Big|=\sqrt{3^2+2^2+6^2}$
$=\sqrt{9+4+36}$
$=\sqrt{49}$
$=7$
$\therefore$ Unit vector in the direction of $\overrightarrow{\text{PQ}}=\frac{\overrightarrow{\text{PQ}}}{\Big|\overrightarrow{\text{PQ}}\Big|}=\frac{3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}}7=\frac{1}7\big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\big)$
View full question & answer→Question 1642 Marks
Write a vector satisfying $\vec{\text{a}}.\hat{\text{i}}=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=1.$
AnswerLet $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$\vec{\text{a}}.\hat{\text{i}}=\text{a}_1$
$\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\text{a}_1+\text{a}_2$
$\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=\text{a}_1+\text{a}_2+\text{a}_3$
Given that
$\vec{\text{a}}.\hat{\text{i}}=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=1$
$\Rightarrow\text{a}_1=\text{a}_1+\text{a}_2=\text{a}_1+\text{a}_2+\text{a}_3=1$
$\Rightarrow\text{a}_1=1;\text{a}_1+\text{a}_2=1;\text{a}_1+\text{a}_2+\text{a}_3=1$
$\Rightarrow\text{a}_1=1;1+\text{a}_2=1;1+\text{a}_2+\text{a}_3=1$
$\Rightarrow\text{a}_1=1;\text{a}_2=0;1+0+\text{a}_3=1$
$\Rightarrow\text{a}_1=1;\text{a}_2=0;\text{a}_3=0$
So, $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}=1\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}=\hat{\text{i}}$
View full question & answer→Question 1652 Marks
If $\vec{\text{a}}.\vec{\text{a}}=0$ and $\vec{\text{a}}.\vec{\text{b}}=0,$ what can you conclude about the vector $\vec{\text{b}}?$
AnswerGiven that $\vec{\text{a}}.\vec{\text{a}}=0$
$\Rightarrow|\vec{\text{a}}|^2=0$
$\Rightarrow|\vec{\text{a}}|=0\dots(1)$
Also, given that
$\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=0$ (where $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$)
$\Rightarrow0\big|\vec{\text{b}}\big|\cos\theta=0$ [From (1)]
$\Rightarrow0=0$
So, it means that for any vector $\vec{\text{b}},$ the given equation $\vec{\text{a}}.\vec{\text{b}}=0$ is satisfid.
View full question & answer→Question 1662 Marks
For given vectors, $\vec{a}=2\hat{i}-\hat{j}+2\hat{k}\ \ \text{and}\ \vec{b}=-\hat{i}+\hat{j}-\hat{k},$ find the unit vector in the direction of the vector $\vec{a}+\vec{b}.$
AnswerThe given vectors are $\vec{a}=2\hat{i}-\hat{j}+2\hat{k}\ \ \text{and}\ \vec{b}=-\hat{i}+\hat{j}-\hat{k}$ $\vec{a}=2\hat{i}-\hat{j}+2\hat{k}$ $\vec{b}=-\hat{i}+\hat{j}-\hat{k}$ $\therefore\vec{a}+\vec{b}=(2-1)\hat{i}+(-1+1)\hat{j}+(2-1)\hat{k}$ $=1\hat{i}+0\hat{j}+1\hat{k}=\hat{i}+\hat{k}$ $\big|\vec{a}+\vec{b}\big|=\sqrt{1^2+1^2}=\sqrt{2}$Hence, the unit vector in the direction of $\big(\vec{a}+\vec{b}\big)$ is
$\frac{\big(\vec{a}+\vec{b}\big)}{\big|\vec{a}+\vec{b}\big|}=\frac{\hat{i}+\hat{k}}{\sqrt{2}}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k}$
View full question & answer→Question 1672 Marks
For any two vectors $\vec{\text{a}}$ and $\vec{\text{b}},$ write when $\big|\vec{\text{a}}+\vec{\text{b}}\big|=\big|\vec{\text{a}}-\vec{\text{b}}\big|$ holds.
AnswerGiven that
$\big|\vec{\text{a}}+\vec{\text{b}}\big|=\big|\vec{\text{a}}-\vec{\text{b}}\big|$
Squaring both sides, we get
$\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=\big|\vec{\text{a}}-\vec{\text{b}}\big|^2$
$\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}.\vec{\text{b}}$
$\Rightarrow4\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\vec{\text{a}}.\vec{\text{b}}=0$
⇒ $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendiculalr.
View full question & answer→Question 1682 Marks
If $\vec{\text{a}},\vec{\text{b}}$ are two vectors, then write the truth value of the following statement:$\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|\Rightarrow\vec{\text{a}}=\vec{\text{b}}$
AnswerFalse
$\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|\Rightarrow\vec{\text{a}}=\vec{\text{b}}$
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 $\vec{\text{a}}\neq\vec{\text{b}}$
View full question & answer→Question 1692 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.Internally
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 internally.
Therefore position vector of $\text{R}=\frac{2\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)+1\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)}{2+1}$
$=\frac{1}3\big(-\hat{\text{i}}+4\hat{\text{j}}+3\hat{\text{k}}\big)$
View full question & answer→Question 1702 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors, find the angle between $\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{a}}-\vec{\text{b}}.$
AnswerWe have
$|\vec{\text{a}}|=1$ and $\big|\vec{\text{b}}\big|=1\dots(1)$
Now,
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2$
$=1^2-1^2$ [using (1)]
$=0$
$\Rightarrow\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{a}}-\vec{\text{b}}$ are perpendicular.
$\therefore$ angle between $\big(\vec{\text{a}}+\vec{\text{b}}\big)$ and $\big(\vec{\text{a}}-\vec{\text{b}}\big)$ is 90°.
View full question & answer→Question 1712 Marks
Write the value of $\big(\vec{\text{a}}.\hat{\text{i}}\big)\hat{\text{i}}+\big(\vec{\text{a}}.\hat{\text{j}}\big)\hat{\text{j}}+\big(\vec{\text{a}}.\hat{\text{k}}\big)\hat{\text{k}},$ where $\vec{\text{a}}$ is any vector.
AnswerLet $\vec{\text{a}}=\text{a}_1\vec{\text{i}}+\text{a}_2\vec{\text{j}}+\text{a}_3\vec{\text{k}}$
Now,
$\big(\vec{\text{a}}.\hat{\text{i}}\big)\hat{\text{i}}+\big(\vec{\text{a}}.\hat{\text{j}}\big)\hat{\text{j}}+\big(\vec{\text{a}}.\hat{\text{k}}\big)\hat{\text{k}}$
$=\text{a}_1\vec{\text{i}}+\text{a}_2\vec{\text{j}}+\text{a}_3\vec{\text{k}}$
$=\vec{\text{a}}$
View full question & answer→Question 1722 Marks
If $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}},$ then write the value of $\big|\vec{\text{r}}\times\hat{\text{i}}\big|^2.$
AnswerGiven: $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
Now,
$\vec{\text{i}}=\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$
$\vec{\text{r}}\times\vec{\text{i}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{x}&\text{y}&\text{z}\\1&0&0 \end{vmatrix}$
$=0\hat{\text{i}}+\text{z}\hat{\text{j}}-\text{y}\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{r}}\times\vec{\text{i}}\big|=\sqrt{\text{x}^2+\text{y}^2}$
$\Rightarrow\big|\vec{\text{r}}\times\vec{\text{i}}\big|^2=\text{x}^2+\text{y}^2$
View full question & answer→Question 1732 Marks
Find the projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ if $\vec{\text{a}}.\vec{\text{b}}=8$ and $\vec{\text{b}}=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}.$
AnswerWe have
$\vec{\text{a}}.\vec{\text{b}}=8$ and $\vec{\text{b}}=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}$
The projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is
$\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{b}}|}$
$=\frac{8}{\sqrt{4+36+9}}$
$=\frac{8}{7}$
View full question & answer→Question 1742 Marks
Find a vector $\vec{\text{a}}$ of magnitude $5\sqrt2$, making an angle of $\frac{\pi}4$ with x-axis, $\frac{\pi}2$ with y-axis and an acute angle $\theta$ with z-axis.
AnswerIt is given that vector $\vec{\text{a}}$ makes an angle of $\frac{\pi}4$ with x-axis, $\frac{\pi}2$ with y-axis and an acute angle $\theta$ with z-axis.$\therefore\ \text{l}=\cos\frac{\pi}4=\frac{1}{\sqrt2},\text{m}=\cos\frac{\pi}2=0,\text{n}=\cos\theta$
Now,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\ \frac{1}2+0+\cos^2\theta=1$
$\Rightarrow\ \cos^2\theta=1-\frac{1}2=\frac{1}2$
$\Rightarrow\ \cos\theta=\frac{1}{\sqrt2}$ ($\theta$ is acute)
We know that
$\vec{\text{a}}=|\vec{\text{a}}|\big(\text{l}\hat{\text{i}}+\text{m}\hat{\text{j}}+\text{n}\hat{\text{k}}\big)$
$\Rightarrow\ \vec{\text{a}}=5\sqrt2\Big(\frac{1}{\sqrt2}\hat{\text{i}}+0\hat{\text{j}}+\frac{1}{\sqrt2}\hat{\text{k}}\Big)$
$\Rightarrow\ \vec{\text{a}}=5\big(\hat{\text{i}}+0\hat{\text{j}}+\hat{\text{k}}\big)$
View full question & answer→Question 1752 Marks
Find $\lambda,$ if $\big(2\hat{\text{i}}+6\hat{\text{j}}+14\hat{\text{k}}\big)\times\big(\hat{\text{i}}-\lambda\hat{\text{j}}+7\hat{\text{k}}\big)=\vec{0}.$
AnswerGiven: $\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&6&14\\1&-\lambda&7\end{vmatrix}=\vec{0}$
$\Rightarrow\hat{\text{i}}(42+14\lambda)-0\hat{\text{j}}+\hat{\text{k}}(-2\lambda-6)=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$
$\Rightarrow42+14\lambda=0;-2\lambda-6=0$
$\Rightarrow\lambda=-3$ (This satisfies the above equations)
View full question & answer→Question 1762 Marks
Write the position vector of a point dividing the line segment joining points A and B with position vectors $\vec{\text{a}}\text{ and }\vec{\text{b}}$ externally in the ratio 1 : 4, where $\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{b}}=-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$.
AnswerThe position vectors of A and B are
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{b}}=-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
Let C divides AB in the ratio such that AB : CB = 1 : 4
Position vector of $\text{C}=\frac{1\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)-4\big(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}\big)}{1-4}$
$=\frac{-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}-8\hat{\text{i}}-12\hat{\text{j}}-16\hat{\text{k}}}{-3}$
$=\frac{-9\hat{\text{i}}-11\hat{\text{j}}-15\hat{\text{k}}}{-3}$
$=3\hat{\text{i}}+\frac{11\hat{\text{j}}}3+5\hat{\text{k}}$
View full question & answer→Question 1772 Marks
Find $\vec{\text{a}}.\vec{\text{b}}$ when
$\vec{\text{a}}=\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{k}}$
Answer$\vec{\text{a}}.\vec{\text{b}}=(\hat{\text{j}}+2\hat{\text{k}}).(2\hat{\text{i}}+\hat{\text{k}})$
$=(0\times\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})(2\hat{\text{i}}+0\times\hat{\text{j}}+\hat{\text{k}})$
$=(0)(2)+(1)(0)+(2)(1)$
$=0+0+2$
$\vec{\text{a}}.\vec{\text{b}}=2$
View full question & answer→Question 1782 Marks
Find the unit vector in the direction of the vector $\vec{a}=\hat{i}+\hat{j}+2\hat{k}.$
AnswerThe unit vector $\hat{a}$ in the direction of vector $\vec{a}=\hat{i}+\hat{j}+2\hat{k}$ is given by $\hat{a}=\frac{\vec{a}}{|a|}.$
$\Big|\vec{a}\Big|=\sqrt{1^2+1^2+2^2}=\sqrt{1+1+4}=\sqrt{6}$
$\therefore{\hat{a}}=\frac{\vec{a}}{\big|\vec{a}\big|}=\frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{6}}=\frac{1}{\sqrt{6}}\hat{i}+\frac{1}{\sqrt{6}}\hat{j}+\frac{2}{\sqrt{6}}\hat{k}$
View full question & answer→Question 1792 Marks
For any two vectore $\vec{\text{a}}$ and $\vec{\text{b}}$, show that $\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=0\Leftrightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|.$
AnswerWe have
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=0$
$\Rightarrow|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2=0$
$\Rightarrow|\vec{\text{a}}|^2=\big|\vec{\text{b}}\big|^2$
$\Rightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|$
View full question & answer→Question 1802 Marks
Find a unit vector in the direction of the vector $\vec{\text{a}}=3\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}$.
AnswerGiven: $\vec{\text{a}}=3\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}$ Then, $|\vec{\text{a}}|=\sqrt{3^2+(-2)^2+6^2}$ $=\sqrt{9+4+36}$ $=\sqrt{49}$ $=7$ $\therefore$ Unit vector $=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{3\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}}7$$=\frac{3}7\hat{\text{i}}-\frac{2}7\hat{\text{j}}+\frac{6}7\hat{\text{k}}$
View full question & answer→Question 1812 Marks
For what value of $\lambda$ are the vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ perpendicular to each other if
$\vec{\text{a}}=\lambda\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=5\hat{\text{i}}-9\hat{\text{j}}+2\hat{\text{k}}$
AnswerIf the vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular to each other, then
$\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\Big(\lambda\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\Big).\Big(5\hat{\text{i}}-9\hat{\text{j}}+2\hat{\text{k}}\Big)=0$
$\Rightarrow5\lambda-18+2=0$
$\Rightarrow5\lambda-16=0$
$\Rightarrow5\lambda=16$
$\Rightarrow\lambda=\frac{16}{5}$
View full question & answer→Question 1822 Marks
What is the angle between vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ with magnitudes 2 and $\sqrt{3}$ respectively? Given $\vec{\text{a}}.\vec{\text{b}}=\sqrt{3}.$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
Given that
$|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=\sqrt{3}$ and $\vec{\text{a}}.\vec{\text{b}}=\sqrt{3}$
We know that
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow\sqrt{3}=(2)(\sqrt{3})\cos\theta$
$\Rightarrow\cos\theta=\frac{\sqrt{3}}{2\sqrt{3}}$
$\Rightarrow\cos\theta=\frac{1}2{}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{1}2{}\big)=\frac{\pi}{3}$
View full question & answer→Question 1832 Marks
If the vectors $3\hat{\text{i}}+\text{m}\hat{\text{j}}+\hat{\text{k}}$ and $2\hat{\text{i}}-\hat{\text{j}}-8\hat{\text{k}}$ are orthonal, find m.
AnswerIt is given that the vectors are othgonal. so, their dot product is zero.
$\big(3\hat{\text{i}}+\text{m}\hat{\text{j}}+\hat{\text{k}}\big).\big(2\hat{\text{i}}-\hat{\text{j}}-8\hat{\text{k}}\big)=0$
$\Rightarrow6-\text{m}-8=0$
$\Rightarrow-\text{m}-2=0$
$\Rightarrow\text{m}=-2$
View full question & answer→Question 1842 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are mutually perpendicular unit vectors, write the value of $\big|\vec{\text{a}}+\vec{\text{b}}\big|.$
Answer$\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors and they are perpendicular.
$\Rightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=1;\vec{\text{a}}.\vec{\text{b}}=0\dots(1)$
Now,
$\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}}$
$=1+1+2(0)$ [using (1)]
$=2$
$\therefore\big|\vec{\text{a}}+\vec{\text{b}}\big|=\sqrt{20}$
View full question & answer→Question 1852 Marks
Evaluate the following:
$\big[\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat{\text{j}}\hat{\text{k}}\hat{\text{i}}\big]+\big[\hat{\text{k}}\hat{\text{i}}\hat{\text{j}}\big]$
AnswerWe have,
$\big[\hat{\text{i}}\hat{\text{ j}}\hat{\text{ k}}\big]+\big[\hat{\text{j}}\hat{\text{ k}}\hat{\text{ i}}\big]+\big[\hat{\text{k}}\hat{\text{i}}\hat{\text{j}}\big]$
$=(\hat{\text{i}}\times\hat{\text{j}}).\hat{\text{k}}(\hat{\text{j}}\times\hat{\text{k}}).\hat{\text{i}}+(\hat{\text{k}}\times\hat{\text{i}}).\hat{\text{j}}$
$=\hat{\text{k}}.\hat{\text{k}}+\hat{\text{i}}.\hat{\text{i}}+\hat{\text{j}}.\hat{\text{j}}$
$=1+1+1$
$=3$
Therefore, $\big[\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat{\text{j}}\hat{\text{k}}\hat{\text{i}}\big]+\big[\hat{\text{k}}\hat{\text{i}}\hat{\text{j}}\big]=3$
View full question & answer→Question 1862 Marks
Can a vector have direction angles 45º, 60º, 120º?
AnswerYes,
Let a vector makes an angle $\alpha=45^{\circ},\ \beta=60^{\circ},\ \gamma=120^{\circ}$ with OX, OY, OZ respectively.
Let l, m, n be the direction cosines of the vector. Then,
$\text{l}=\cos45^{\circ}=\frac{1}{\sqrt{2}}$
$\text{m}=\cos60^{\circ}=\frac{1}2$
$\text{n}=\cos120^{\circ}=-\frac{1}2$
So,
$\text{l}^2+\text{m}^2+\text{n}^2$
$=\frac{1}2+\frac{1}4+\frac{1}4$
$=1$
Since, the vector has direction cosines such that $\text{l}^2+\text{m}^2+\text{n}^2=1$
Hence, a vector can have direction angles 45º, 60º, 120º
View full question & answer→Question 1872 Marks
If $\overrightarrow{\text{AO}}+\overrightarrow{\text{OB}}=\overrightarrow{\text{BO}}+\overrightarrow{\text{OC}}$, prove that A, B, C are collinear points.
AnswerHere, $\overrightarrow{\text{AO}}+\overrightarrow{\text{OB}}=\overrightarrow{\text{BO}}+\overrightarrow{\text{OC}}$
$\overrightarrow{\text{OA}}-\overrightarrow{\text{BO}}=\overrightarrow{\text{OB}}-\overrightarrow{\text{CO}}$
$\overrightarrow{\text{AB}}=\overrightarrow{\text{BC}}$
So, $\overrightarrow{\text{AB}}$ is parallel to $\overrightarrow{\text{BC}}$ but $\vec{\text{B}}$ is a common vector. Hence,
A, B, C are collinear.
View full question & answer→Question 1882 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors such that $\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big).=0,$ find the relation betwen the magnitudes of $\vec{\text{a}}$ and $\vec{\text{b}}.$
AnswerGiven that$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=0$
$\Rightarrow|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2=0$
$\Rightarrow|\vec{\text{a}}|^2=\big|\vec{\text{b}}\big|^2$
$\therefore|\vec{\text{a}}|=\big|\vec{\text{b}}\big|$
View full question & answer→Question 1892 Marks
Find the angle at which the following vectors are inclined to each of the coordinate axes:
$\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}}=\hat{\text{j}}-\hat{\text{k}}$ are proportional to 0, 1, -1. Therefore,
Direction cosine of $\vec{\text{r}}$ are $\frac{0}{\sqrt{0+1^2+(-1^2)}},\frac{1}{\sqrt{0+1^2+(-1^2)}},\frac{-1}{\sqrt{0+1^2+(-1^2)}}$ or $0,\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}}$
$\therefore\cos\alpha=0,\cos\beta=\frac{1}{\sqrt{2}},\cos\gamma=\frac{-1}{\sqrt{2}}$
$\Rightarrow\alpha=\cos^{-1}=(0),\beta=\cos^{-1}=\Big(\frac{-1}{\sqrt{2}}\Big),\gamma=\cos^{-1}=\Big(\frac{-1}{\sqrt{2}}\Big)$
$\Rightarrow\alpha=\frac{\pi}{2},\beta=\frac{\pi}{4},\gamma=\frac{3\pi}{4}$
View full question & answer→Question 1902 Marks
Find the direction cosines of the vector joining the points A (1, 2, -3) and B (-1, -2, 1), directed from A to B.
AnswerThe given points are A (1, 2 -3) and B (-1, -2, 1).
$\therefore{\overrightarrow{\text{AB}}}=(-1-1)\hat{i}+(-2-2)\hat{j}+\{1-(-3)\}\hat{k}$
$\Rightarrow\overrightarrow{\text{AB}}=-2\hat{i}-4\hat{j}+4\hat{k}$
$\therefore\bigg|\overrightarrow{\text{AB}}\bigg|=\sqrt{(-2)^2+(-4)^2+4^2}=\sqrt{4+16+16}=\sqrt{36}=6$
Hence, the direction cosines of $\overrightarrow{\text{AB}}\ \text{are}\ \bigg(-\frac{2}{6},-\frac{4}{6},\frac{4}{6}\bigg)=\bigg(-\frac{1}{3},\frac{2}{3},\frac{2}{3}\bigg).$
View full question & answer→Question 1912 Marks
For any two vectors $\vec{\text{a}} $ and $\vec{\text{b}},$ write when $\big|\vec{\text{a}}+\vec{\text{b}}\big|=|\vec{\text{a}}|+\big|\vec{\text{b}}\big|$ holds.
AnswerGiven that $\big|\vec{\text{a}}+\vec{\text{b}}\big|=|\vec{\text{a}}|+\big|\vec{\text{b}}\big|$ Squaring both sides, we get $\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=\big(|\vec{\text{a}}|+\big|\vec{\text{b}}\big|\big)^2$ $\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2|\vec{\text{a}}|\big|\vec{\text{b}}\big|$ $\Rightarrow\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|$ $\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=|\vec{\text{a}}|\big|\vec{\text{b}}\big|$ (where $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$) $\Rightarrow\cos\theta=1$ $\Rightarrow\theta=0^{\circ}$$\Rightarrow\vec{\text{a}}$ and $\vec{\text{b}}$ are parallel.
View full question & answer→Question 1922 Marks
Write a vector in the direction of vector $5\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$ which has magnitude of 8 unit.
AnswerGiven:
$\vec{\text{a}}=5\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$|\vec{\text{a}}|=\sqrt{5^2+(-1)^2+2^2}$
$=\sqrt{25+1+4}$
$=\sqrt{30}$
$\therefore$ Position vector in the direction of vector $=8\times\frac{\vec{\text{a}}}{|\vec{\text{a}}|}$
$=\frac{8}{\sqrt{30}}\big(5\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big)$
View full question & answer→Question 1932 Marks
Find $\vec{\text{a}}.\vec{\text{b}}$ when
$\vec{\text{a}}=\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}}$
Answer$\vec{\text{a}}.\vec{\text{b}}=(\hat{\text{j}}-\hat{\text{k}}).(2\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}})$
$=(0\times\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})(2\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}})$
$=(0)(2)+(1)(3)+(-1)(-2)$
$=0+3+2$
$\vec{\text{a}}.\vec{\text{b}}=5$
View full question & answer→Question 1942 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are position vectors of the points A, B and C respectively, write the value of$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{AC}}$.
AnswerGiven: $\vec{\text{a}},\vec{\text{b}}\text{ and }\vec{\text{c}}$ are the position vectors of A, B, C respectively. Then,
$\overrightarrow{\text{AB}}=\vec{\text{b}}-\vec{\text{a}}$
$\overrightarrow{\text{BC}}=\vec{\text{c}}-\vec{\text{b}}$
$\overrightarrow{\text{AC}}=\vec{\text{c}}-\vec{\text{a}}$
Therefore,
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{AC}}\\=\vec{\text{b}}-\vec{\text{a}}+\vec{\text{c}}-\vec{\text{b}}+\vec{\text{c}}-\vec{\text{a}}$
$=2\big(\vec{\text{c}}-\vec{\text{a}}\big)$
View full question & answer→Question 1952 Marks
Write the value of $\big[\hat{\text{i}}+\hat{\text{j}}\hat{\text{j}}+\hat{\text{k}}\hat{\text{k}}+\hat{\text{i}}\big].$
AnswerWe have
$\big[\hat{\text{i}}+\hat{\text{j}}\hat{\text{j}}+\hat{\text{k}}\hat{\text{k}}+\hat{\text{i}}\big]=\big[\big(\hat{\text{i}}+\hat{\text{j}}\big)\times\big(\hat{\text{j}}+\hat{\text{k}}\big)\big].\big(\hat{\text{k}}+\hat{\text{i}}\big)$ $\big(\therefore\big[\vec{\text{a}}\vec{\text{ b }}\vec{\text{c}}\big]=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big)$
$=\big[\hat{\text{i}}\times\hat{\text{j}}+\hat{\text{i}}\times\hat{\text{k}}+\hat{\text{j}}\times\hat{\text{j}}+\hat{\text{j}}\times\hat{\text{k}}\big].\big(\hat{\text{k}}+\hat{\text{i}}\big)$
$=\big[\hat{\text{k}}-\hat{\text{j}}+\hat{\text{i}}\big].\big(\hat{\text{k}}+\hat{\text{i}}\big)$
$=\big[\big(\hat{\text{k}}.\hat{\text{k}}\big)+\big(\hat{\text{k}}.\hat{\text{i}}\big)-\big(\hat{\text{j}}.\hat{\text{k}}\big)-\big(\hat{\text{j}}.\hat{\text{i}}\big)+\big(\hat{\text{i}}.\hat{\text{k}}\big)+\big(\hat{\text{i}}.\hat{\text{i}}\big)\big]$
$=1+0-0-0+0+1=2$
View full question & answer→Question 1962 Marks
Find the projection of the vector $\hat{i}-\hat{j}$ on the vector $\hat{i}+\hat{j}.$
Answer$\text{Let}\ \ \vec{a}=\hat{i}-\hat{j}=\hat{i}-\hat{j}+0\hat {k}\ \text{and}\ \vec{b}=\hat{i}+\hat{j}$ $=\hat{i}+\hat{j}+0\hat {k}$
Projection of vector $\vec{a}\ \text{and}\ \vec{b}=\frac{\vec{a}.\vec{b}}{\big|\vec{b}\big|}=\frac{(1)(1)+(-1)(1)+0(0)}{\sqrt{(1)^2+(1)^2+(0)^2}}$ $=\frac{1-1+0}{\sqrt{2}}=\frac{0}{\sqrt{2}}=0$
If projection of vector $\vec{a}\ \text{and}\ \vec{b}$ is zero, then vector $\vec{a}$ is perpendicular to $\vec{b}.$
View full question & answer→Question 1972 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 1982 Marks
Find the unit vector in the direction of $3\hat{\text{i}}+4\hat{\text{j}}-12\hat{\text{k}}$.
AnswerLet $\vec{\text{a}}=3\hat{\text{i}}+4\hat{\text{j}}-12\hat{\text{k}}$Then, $\big|\vec{\text{a}}\big|=\sqrt{3^2+4^2+(-12)^2}$
$=\sqrt{9+16+144}$
$=\sqrt{169}$
$=13$
So, a unit vector in the direction of $\vec{\text{a}}$ is given by
$\hat{\text{a}}=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{1}{13}\big(3\hat{\text{i}}+4\hat{\text{j}}-12\hat{\text{k}}\big)$
$=\frac{3}{13}\hat{\text{i}}+\frac{4}{13}\hat{\text{j}}-\frac{12}{13}\hat{\text{k}}$
View full question & answer→Question 1992 Marks
Write a unit vector in the direction of $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$.
AnswerGiven:$\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
$\big|\vec{\text{b}}\big|=\sqrt{2^2+1^2+2^2}$ $=\sqrt{4+1+4}$ $=\sqrt9$ $=3$ $\therefore$ Unit vector $=\frac{\vec{\text{b}}}{\big|\vec{\text{b}}\big|}=\frac{1}3\big(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$ $=\frac{2}3\hat{\text{i}}+\frac{1}3\hat{\text{j}}+\frac{2}3\hat{\text{k}}$
View full question & answer→Question 2002 Marks
Find the angle between two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ with magnitudes 1 and 2 respectively and when $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{3.}$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
We know $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$\Rightarrow\sqrt{3}=(1)(2)\sin\theta$
$\Rightarrow\sin\theta=\frac{\sqrt{3}}{2}$
$\Rightarrow\theta=\frac{\pi}{3}$
View full question & answer→