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Maths 2 Marks

VECTOR ALGEBRA · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 215 questions.

Questions · Page 2 of 5

2 Marks

Question 512 Marks
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 $2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}$.
Answer
We have, $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$and $\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
$\therefore\ 2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}=2\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)\\-\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)+3\big(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)$
$=3\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$
A unit vector parallel to $2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}$ is given by $\frac{2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}}{\big|2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}\big|}=\frac{(3\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}})}{\sqrt{3^2+(-3)^2+2^2}}$
$=\frac{(3\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}})}{\sqrt{22}}$
$=\frac{3}{\sqrt{22}}\hat{\text{i}}-\frac{3}{\sqrt{22}}\hat{\text{j}}+\frac{2}{\sqrt{22}}\hat{\text{k}}$
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Question 522 Marks
Write the value of $\hat{\text{i}}.\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\hat{\text{j}}.\big(\hat{\text{k}}\times\hat{\text{i}}\big)+\hat{\text{k}}.\big(\hat{\text{j}}\times\hat{\text{i}}\big).$
Answer
$\hat{\text{i}}.\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\hat{\text{j}}.\big(\hat{\text{k}}\times\hat{\text{i}}\big)+\hat{\text{k}}.\big(\hat{\text{j}}\times\hat{\text{i}}\big)$

$=\hat{\text{i}}.\hat{\text{i}}+\hat{\text{j}}.\hat{\text{j}}+\hat{\text{k}}\big(-\hat{\text{k}}\big)$

$=|\hat{\text{i}}|^2+|\hat{\text{j}}|^2-|\hat{\text{k}}|^2$

$=1+1-1$ $\big(\because|\hat{\text{i}}|=1,|\hat{\text{j}}|=1\text{ and } |\hat{\text{k}}|=1\big)$

$=1$

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Question 532 Marks
Write the value of $\hat{\text{i}}\times\big(\hat{\text{j}}\times\hat{\text{k}}\big).$
Answer
$\hat{\text{i}}\times\big(\hat{\text{j}}\times\hat{\text{k}}\big)$

$=\hat{\text{i}}\times\hat{\text{i}}$

$=0$

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Question 542 Marks
Write a vector of magnitude 12 units which makes 45º angle with x-axis, 60º angle with y-axis and an obtuse angle with z-axis.
Answer
Suppose a vector $\vec{\text{r}}$ makes an angle 45º with OX, 60º with OY  and having magnitude 12 units. $\text{l}=\cos45^{\circ}=\frac{1}{\sqrt2}$ and $\text{m}=\cos60^{\circ}=\frac{1}2$

Now, $\text{l}^2+\text{m}^2+\text{n}^2=1$

$\Rightarrow\ \frac{1}2+\frac{1}4+\text{n}^2=1$

$\Rightarrow\ \text{n}^2=\frac{1}4$

$\Rightarrow\ =-\frac{1}2$ $[\because$ The angle with the z-axis is obtuse$]$

Therefore,

$\vec{\text{r}}=|\vec{\text{r}}|\big(\text{l}\hat{\text{i}}+\text{m}\hat{\text{j}}+\text{n}\hat{\text{k}}\big)$

$=12\Big(\frac{1}{\sqrt2}\hat{\text{i}}+\frac{1}2\hat{\text{j}}-\frac{1}2\hat{\text{k}}\Big)$

$=6\big(\sqrt2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$

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Question 552 Marks
If G denots the centroid of $\triangle\text{ABC}$, then write the value of $\overrightarrow{\text{GA}}+\overrightarrow{\text{GB}}+\overrightarrow{\text{GC}}$.
Answer
Let $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ be the position vectors of the vertices A, B, C respectively. Then, the position vector of the centroid G is $\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3$
Thus,
$\overrightarrow{\text{GA}}+\overrightarrow{\text{GB}}+\overrightarrow{\text{GC}}$
$=\vec{\text{a}}-\Big(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3\Big)+\vec{\text{b}}-\Big(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3\Big)+\vec{\text{c}}-\Big(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3\Big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)-3\Big(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3\Big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)-\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)$
$=\vec0$
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Question 562 Marks
Write the angle between two vectors ​​$​​\vec{\text{a}}$ and $​​\vec{\text{b}}$ with magnitudes $\sqrt{3}$ and 2 respectively if $​​\vec{\text{a}}​​.\vec{\text{b}}=\sqrt{6}.$
Answer
Let $\theta$ be the angle between$​​\vec{\text{a}}$ and$​​\vec{\text{b}}.$
Given,
$|\vec{\text{a}}|=\sqrt{3};\big|\vec{\text{b}}\big|=2;​​\vec{\text{a}}.\vec{\text{b}}=\sqrt{6}$
We know that
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow\sqrt{6}=(\sqrt{3})(2)\cos\theta$
$\Rightarrow\cos\theta=\frac{\sqrt{6}}{2\sqrt{3}}=\frac{1}{\sqrt{2}}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{1}{\sqrt{2}}\Big)=\frac{\pi}{4}$
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Question 572 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors, then write the value of $\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2.$
Answer
It is given that $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors.

$\Rightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=1\dots(1)$

Now,

$\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2$

$=\big(|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\big)^2+\big(|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\big)^2$

$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2(\cos^2\theta+\sin^2\theta)$

$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$ (1)

$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$

$=1^21^2$ [From (1)]

$=1$

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Question 582 Marks
Write the value $\big(\hat{\text{i}}\times\hat{\text{j}}\big).\hat{\text{k}}+\hat{\text{i}}.\hat{\text{j}}$
Answer
$\big(\hat{\text{i}}\times\hat{\text{j}}\big).\hat{\text{k}}+\hat{\text{i}}.\hat{\text{j}}$

$=\hat{\text{k}}.\hat{\text{k}}+0$

$=|\hat{\text{k}}|^2+0$

$=1^2+0$ $\big(\because|\text{k}|=1\big)$

$=1$

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Question 592 Marks
If $\vec{\text{a}},\vec{\text{b}}$ are two vectors, then write the truth value of the following statement:

$\vec{\text{a}}=-\vec{\text{b}}\Rightarrow\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|$

Answer
True
$\vec{\text{a}}=-\vec{\text{b}}$
Take modulus both sides
$\big|\vec{\text{a}}\big|=\big|-\vec{\text{b}}\big|$
$\Rightarrow\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|$ $\Big[\therefore\ \big|-\vec{\text{b}}\big|=\big|\vec{\text{b}}\big|\Big]$
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Question 602 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 $\theta$.
Answer
A 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.

$\therefore\ \theta=30^{\circ}$

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Question 612 Marks
Write a vector satisfying $\vec{\text{a}}.\hat{\text{i}}=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=1.$
Answer
Let $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$\vec{\text{a}}.\hat{\text{i}}=\text{a}_1$
$\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\text{a}_1+\text{a}_2$
$\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=\text{a}_1+\text{a}_2+\text{a}_3$
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, $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}=1\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}=\hat{\text{i}}$
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Question 622 Marks
If $\vec{\text{a}},\vec{\text{b}}$ are two vectors, then write the truth value of the following statement:

$\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|\Rightarrow\vec{\text{a}}=\vec{\text{b}}$

Answer
False
$\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|\Rightarrow\vec{\text{a}}=\vec{\text{b}}$
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 $\vec{\text{a}}\neq\vec{\text{b}}$
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Question 632 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

Answer
Given: 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 $\text{R}=\frac{2\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)+1\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)}{2+1}$
$=\frac{1}3\big(-\hat{\text{i}}+4\hat{\text{j}}+3\hat{\text{k}}\big)$
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Question 642 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.
Answer
It 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

$\vec{\text{a}}=|\vec{\text{a}}|\big(\text{l}\hat{\text{i}}+\text{m}\hat{\text{j}}+\text{n}\hat{\text{k}}\big)$

$\Rightarrow\ \vec{\text{a}}=5\sqrt2\Big(\frac{1}{\sqrt2}\hat{\text{i}}+0\hat{\text{j}}+\frac{1}{\sqrt2}\hat{\text{k}}\Big)$

$\Rightarrow\ \vec{\text{a}}=5\big(\hat{\text{i}}+0\hat{\text{j}}+\hat{\text{k}}\big)$

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Question 652 Marks
Find $\lambda,$ if $\big(2\hat{\text{i}}+6\hat{\text{j}}+14\hat{\text{k}}\big)\times\big(\hat{\text{i}}-\lambda\hat{\text{j}}+7\hat{\text{k}}\big)=\vec{0}.$
Answer
Given: $\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&6&14\\1&-\lambda&7\end{vmatrix}=\vec{0}$
$\Rightarrow\hat{\text{i}}(42+14\lambda)-0\hat{\text{j}}+\hat{\text{k}}(-2\lambda-6)=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$
$\Rightarrow42+14\lambda=0;-2\lambda-6=0$
$\Rightarrow\lambda=-3$ (This satisfies the above equations)
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Question 662 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors, find the angle between $\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{a}}-\vec{\text{b}}.$
Answer
We 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°.
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Question 672 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.
Answer
Given: $\vec{\text{a}}\text{ and }\vec{\text{b}}$ 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 $\frac{1}3\big(\vec{\text{a}}+2\vec{\text{b}}\big)$

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Question 682 Marks
For any two vectore $\vec{\text{a}}$ and $\vec{\text{b}}$, show that $\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=0\Leftrightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|.$
Answer
We have 
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=0$
$\Rightarrow|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2=0$
$\Rightarrow|\vec{\text{a}}|^2=\big|\vec{\text{b}}\big|^2$
$\Rightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|$
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Question 692 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.
Answer
P(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 $\overrightarrow{\text{PQ}}=\frac{\overrightarrow{\text{PQ}}}{\Big|\overrightarrow{\text{PQ}}\Big|}=\frac{3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}}7=\frac{1}7\big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\big)$
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Question 702 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.
Answer
It is given that the vectors are othgonal. so, their dot product is zero.
$\big(3\hat{\text{i}}+\text{m}\hat{\text{j}}+\hat{\text{k}}\big).\big(2\hat{\text{i}}-\hat{\text{j}}-8\hat{\text{k}}\big)=0$
$\Rightarrow6-\text{m}-8=0$
$\Rightarrow-\text{m}-2=0$
$\Rightarrow\text{m}=-2$
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Question 712 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.
Answer
Given:
$\vec{\text{a}}=5\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$|\vec{\text{a}}|=\sqrt{5^2+(-1)^2+2^2}$
$=\sqrt{25+1+4}$
$=\sqrt{30}$
$\therefore$ Position vector in the direction of vector $=8\times\frac{\vec{\text{a}}}{|\vec{\text{a}}|}$
$=\frac{8}{\sqrt{30}}\big(5\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big)$
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Question 722 Marks
Write a unit vector in the direction of $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$.
Answer
Given:

 $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$

$\big|\vec{\text{b}}\big|=\sqrt{2^2+1^2+2^2}$

$=\sqrt{4+1+4}$

$=\sqrt9$

$=3$

$\therefore$ Unit vector $=\frac{\vec{\text{b}}}{\big|\vec{\text{b}}\big|}=\frac{1}3\big(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$

$=\frac{2}3\hat{\text{i}}+\frac{1}3\hat{\text{j}}+\frac{2}3\hat{\text{k}}$

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Question 732 Marks
Find the angle between two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ with magnitudes 1 and 2 respectively and when $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{3.}$
Answer
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
We know $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$\Rightarrow\sqrt{3}=(1)(2)\sin\theta$
$\Rightarrow\sin\theta=\frac{\sqrt{3}}{2}$
$\Rightarrow\theta=\frac{\pi}{3}$
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Question 742 Marks
If  $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},$ find the magnitude of $\vec{\text{a}}\times\vec{\text{b}}.$
Answer
Given:

$\vec{\text{a}}=2\hat{\text{i}}+0\hat{\text{j}}+\hat{\text{k}}$

$\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$

$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&0&1\\1&1&1 \end{vmatrix}$

$=(0-1)\hat{\text{i}}-(2-1)\hat{\text{j}}+(2-0)\hat{\text{k}}$

$=-\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$

$\Rightarrow\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{(-1)^2+(1-)^2+2^2}$

$=\sqrt{6}$

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Question 752 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$​​​​\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{AC}}$.
Answer
Given: $\vec{\text{a}},\vec{\text{b}}\text{ and }\vec{\text{c}}$ 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,
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{AC}}\\=\vec{\text{b}}-\vec{\text{a}}+\vec{\text{c}}-\vec{\text{b}}+\vec{\text{c}}-\vec{\text{a}}$
$=2\big(\vec{\text{c}}-\vec{\text{a}}\big)$
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Question 762 Marks
Prove that 1, 1, 1 cannot be direction cosines of a straight line.
Answer
Let 1, 1, 1 be the direction cosines of a straight line. Then,
$1^2+1^2+1^2=3\neq1$
Since direction cosines of a line which makes equal angle with the axes must satisfy
$\text{l}^2+\text{m}^2+\text{n}^2=1$
Hence 1, 1, 1 cannot be the direction cosines of a straight line.
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Question 772 Marks
Find the angle betwwen two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ if

$|\vec{\text{a}}|=\sqrt{3},\big|\vec{\text{b}}\big|=2$ and $\vec{\text{a}}.\vec{\text{b}}=\sqrt{6}$

Answer
We have,
$|\vec{\text{a}}|=\sqrt{3},\big|\vec{\text{b}}\big|=2$ and $\vec{\text{a}}.\vec{\text{b}}=\sqrt{6}$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$. then
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|}$
$=\frac{\sqrt{6}}{\sqrt{3}\times2}$
$=\frac{1}{\sqrt{2}}$
$\theta=\cos^{-1}\Big(\frac{1}{\sqrt{2}}\Big)$
$=\frac{\pi}{4}$
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Question 782 Marks
What is the cosine of the angle with the vector $\sqrt2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ makes with y-axis?
Answer
Given $\sqrt2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
Therefore, direction cosines are $\frac{\sqrt2}{\sqrt{(\sqrt2)^2+1^2+1^2}},\frac{1}{\sqrt{(\sqrt2)^2+1^2+1^2}},\frac{1}{\sqrt{(\sqrt2)^2+1^2+1^2}}$ or $\frac{1}{\sqrt2},\frac{1}2,\frac{1}2$
So, cosine angle with respect to y-axis is $\frac{1}2$
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Question 792 Marks
For what of $\lambda$ are the vectors $\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$ perpendicular to each other?
Answer
We have
$\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
Given that $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular.
$\Rightarrow\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\big(2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}\big).\big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big)=0$
$\Rightarrow2-2\lambda+3=0$
$\Rightarrow5-2\lambda=0$
$\therefore\lambda=\frac{5}{2}$
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Question 802 Marks
Write the angle between the vectors $\vec{\text{a}}\times\vec{\text{b}}$ and $\vec{\text{b}}\times\vec{\text{a}}.$
Answer
$\vec{\text{b}}\times\vec{\text{a}}=-\vec{\text{a}}\times\vec{\text{b}}$
So, $\vec{\text{a}}\times\vec{\text{b}}$ and $\vec{\text{b}}\times\vec{\text{a}}$ are vectors of same magnitude but opposite in direction.
Thus, the angle between the vectors $\vec{\text{a}}\times\vec{\text{b}}$ and $\vec{\text{b}}\times\vec{\text{a}}$ is $180^\circ.$
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Question 812 Marks
Show that the vector $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ is equally inclined with the axes OX, OY and OZ. 
Answer
Let $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
 Then,
$|\vec{\text{a}}|=\sqrt{1^1+1^1+1^1}=\sqrt{3}$
Therefore, the direction ratios of $\vec{\text{a}}$ are $\Big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\Big).$
Now, let $\alpha,\beta$ and $\gamma$ be the angles format by $\vec{\text{a}}$ with the positive directions of x, y and z axes.
Then, we have $\cos\alpha=\frac{1}{\sqrt{3}},\cos\beta=\frac{1}{\sqrt{3}},\cos\gamma=\frac{1}{\sqrt{3}}.$
Hence, the given vector is equally inclined to axes OX, OY and OZ.
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Question 822 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ represent the sides of a triangle taken in order, then write the value of $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$.
Answer
Let ABC be a triangle such that $\overrightarrow{\text{BC}}=\vec{\text{a}},\ \overrightarrow{\text{CA}}=\vec{\text{b}},\ \overrightarrow{\text{AB}}=\vec{\text{c}}$. Then,

$\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}+\overrightarrow{\text{AB}}$

$=\overrightarrow{\text{BA}}+\overrightarrow{\text{AB}}$ $\Big[\because \overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=\overrightarrow{\text{BA}}\Big]$

$=\vec0$

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Question 832 Marks
If P, Q and R are three collinear points such that $\overrightarrow{\text{PQ}}=\vec{\text{a}}\text{ and }\overrightarrow{\text{QR}}=\vec{\text{b}}$. Find the vector $\overrightarrow{\text{PR}}$.
Answer
Given that, P, Q, R are collinear.

It also given that, $\overrightarrow{\text{PQ}}=\vec{\text{a}}\text{ and }\overrightarrow{\text{QR}}=\vec{\text{b}}$

$\overrightarrow{\text{PR}}=\overrightarrow{\text{PQ}}+\overrightarrow{\text{QR}}$

$=\vec{\text{a}}+\vec{\text{b}}$

$\overrightarrow{\text{PR}}=\vec{\text{a}}+\vec{\text{b}}$

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Question 842 Marks
If $\vec{\text{a}}$ is a unit vector, then find $|\vec{\text{x}}|$ in each of the following.

$\big(\vec{\text{x}}-\vec{\text{a}}\big).\big(\vec{\text{x}}+\vec{\text{a}}\big)=12$

Answer
Given that $\vec{\text{a}}$ is a unit vector.

$\big(\vec{\text{x}}-\vec{\text{a}}\big).\big(\vec{\text{x}}+\vec{\text{a}}\big)=12$

$\Rightarrow|\vec{\text{x}}|^2-|\vec{\text{a}}|^2=12$

$\Rightarrow|\vec{\text{x}}|^2-1^2=12$ [From (1)]

$\Rightarrow|\vec{\text{x}}|^2=13$

$\Rightarrow|\vec{\text{x}}|=\sqrt{13}$

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Question 852 Marks
If $\vec{\text{a}}$ is a unit vector such that $\vec{\text{a}}\times\hat{\text{i}}=\hat{\text{j}},$ find $\vec{\text{a}}.\hat{\text{i}}.$
Answer
We know
$\hat{\text{k}}\times\hat{\text{i}}=\hat{\text{j}}\dots(1)$
Given: $\vec{\text{a}}\times\hat{\text{i}}=\hat{\text{j}}\dots(2)$
Comparing (1) and (2), we get
$\vec{\text{a}}=\hat{\text{k}}$
Now,
$\vec{\text{a}}.\hat{\text{i}}=\hat{\text{k}}.\hat{\text{i}}$
$=0$
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Question 862 Marks
Write the value of $\hat{\text{i}}\times\big(\hat{\text{j}}+\hat{\text{k}}\big)+\hat{\text{j}}\times\big(\hat{\text{k}}+\hat{\text{i}}\big)+\hat{\text{k}}\times\big(\hat{\text{i}}+\hat{\text{j}}\big).$
Answer
$\hat{\text{i}}\times\big(\hat{\text{j}}+\hat{\text{k}}\big)+\hat{\text{j}}\times\big(\hat{\text{k}}+\hat{\text{i}}\big)+\hat{\text{k}}\times\big(\hat{\text{i}}+\hat{\text{j}}\big)$
$=\big(\hat{\text{i}}\times\hat{\text{j}}\big)+\big(\hat{\text{i}}\times\hat{\text{k}}\big)+\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\big(\hat{\text{j}}\times\hat{\text{i}}\big)+\big(\hat{\text{k}}\times\hat{\text{i}}\big)+\big(\hat{\text{k}}\times\hat{\text{j}}\big)$
$=\hat{\text{k}}-\hat{\text{j}}+\hat{\text{i}}-\hat{\text{k}}+\hat{\text{j}}-\hat{\text{i}}=\vec{0}$
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Question 872 Marks
If $|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=5$ and $\vec{\text{a}}.\vec{\text{b}}=2,$ find $\big|\hat{\text{a}}-\hat{\text{b}}\big|.$
Answer
We have 

$|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=5$ and $\vec{\text{a}}.\vec{\text{b}}=2$

Now,

$\big|\hat{\text{a}}-\hat{\text{b}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}.\vec{\text{b}}$

$=2^2+5^2-2(2)$

$=4+25-4$

$=25$

$\therefore\big|\vec{\text{a}}-\vec{\text{b}}\big|=\sqrt{25}=5$

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Question 882 Marks
Find the position vector of a point R which divides the line segment joining points $\text{P}\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)$ and $\text{Q}\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$ in the ratio 2 : 1.

Externally

Answer
Given: R divides the line segment joining the points $\text{P}\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big),\text{Q}\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$ in the ratio 2 : 1 externally.
Therefore position vector of $\text{R}=\frac{2\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)+1\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)}{2+1}$
$=-3\hat{\text{i}}+\hat{\text{k}}$
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Question 892 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 $\vec{\text{b}} =4\hat{\text{i}}-\hat{\text{j}}+8\hat{\text{k}}$

Answer
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.
$\big|\vec{\text{a}}\big|=\sqrt{(3)^2+(-2)^2+(-6)^{2}}=\sqrt{49}=7$
$\big|\vec{\text{b}}\big|=\sqrt{(4)^2+(1)^2+(8)^{2}}=\sqrt{81}=9$
$\vec{\text{a}}.\vec{\text{b}}=12+2-48=-34$
$\cos\theta=\frac{\vec{\text{a }}.\vec{\text{ b}}}{\big|\vec{a}\big|\big|\vec{b}\big|}=\frac{-34}{(7)(9)}=\frac{-34}{63}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{-34}{63}\big)$
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Question 902 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 $\vec{\text{r}}$.
Answer
Here, $\alpha=\beta=\gamma$
$\Rightarrow\ \cos\alpha=\cos\beta=\cos\gamma$
$\Rightarrow\ \text{l}=\text{m}=\text{n}=\text{x}$ (say)
We know that,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\text{x}^2+\text{x}^2+\text{x}^2=1$
$3\text{x}^2=1$
$\text{x}^2=\frac{1}3$
$\text{x}=\pm\frac{1}{\sqrt3}$
$\text{l}=\pm\frac{1}{\sqrt3},\ \text{m}=\pm\frac{1}{\sqrt3},\ \text{n}=\pm\frac{1}{\sqrt3}$
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Question 912 Marks
Write the projection of the vector $\hat{\text{i}}+3\hat{\text{j}}+7\hat{\text{k}}$ on the vector $2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}.$
Answer
We 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 $\vec{\text{b}}$
$=\frac{\big(\hat{\text{i}}+3\hat{\text{j}}+7\vec{\text{k}}\big).\big(2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}.\big)}{\big|2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}.\big|}$
$=\frac{1\times2+3\times(-3)+7\times6}{\sqrt{2^2+(-3)^2+6^2}}$
$=\frac{2-9+42}{\sqrt{49}}$
$=\frac{35}{7}$
$=5$
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Question 922 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.
Answer
Let 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}$
$= \frac{\hat{\text{i}} - 5\hat{\text{j}} + 12\hat{\text{k}}}{-1}$
$= -\hat{\text{i}} + 5\hat{\text{j}} - 12\hat{\text{k}}$
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Question 932 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.
Answer
Let $\vec{\text{a}}=\text{a}_1\vec{\text{i}}+\text{a}_2\vec{\text{j}}+\text{a}_3\vec{\text{k}}$
Now,
$\big(\vec{\text{a}}.\hat{\text{i}}\big)\hat{\text{i}}+\big(\vec{\text{a}}.\hat{\text{j}}\big)\hat{\text{j}}+\big(\vec{\text{a}}.\hat{\text{k}}\big)\hat{\text{k}}$
$=\text{a}_1\vec{\text{i}}+\text{a}_2\vec{\text{j}}+\text{a}_3\vec{\text{k}}$
$=\vec{\text{a}}$
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Question 942 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 $\big|\vec{\text{r}}\times\hat{\text{i}}\big|^2.$
Answer
Given: $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
Now,
$\vec{\text{i}}=\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$
$\vec{\text{r}}\times\vec{\text{i}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{x}&\text{y}&\text{z}\\1&0&0 \end{vmatrix}$
$=0\hat{\text{i}}+\text{z}\hat{\text{j}}-\text{y}\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{r}}\times\vec{\text{i}}\big|=\sqrt{\text{x}^2+\text{y}^2}$
$\Rightarrow\big|\vec{\text{r}}\times\vec{\text{i}}\big|^2=\text{x}^2+\text{y}^2$
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Question 952 Marks
For what value of 'a' the vectors $2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$ and $\text{a}\hat{\text{i}}+6\hat{\text{j}}-8\hat{\text{k}}$ are collinear?
Answer
Given: Two vectors, let $\vec{\text{p}}=2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{q}}=\text{a}\hat{\text{i}}+6\hat{\text{j}}-8\hat{\text{k}}$
Since the given vectors are collinear, we have,
$\vec{\text{p}}=\lambda\vec{\text{q}}$
$\Rightarrow\ 2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}=\lambda\big(\text{a}\hat{\text{i}}+6\hat{\text{j}}-8\hat{\text{k}}\big)$
$\Rightarrow\ 2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}=\text{a}\lambda\hat{\text{i}}+6\lambda\hat{\text{j}}-8\lambda\hat{\text{k}}$
$\Rightarrow\ \lambda\text{a}=2,6\lambda=-3$ and $-8\lambda=4$
$\Rightarrow\ \lambda=-\frac{1}2$ and $\text{a}= -4$
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Question 962 Marks
For any two vectors $\vec{\text{a}}$ and $\vec{\text{b}},$ find $\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{b}}.$
Answer
Let:

$\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$

$\vec{\text{b}}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$

$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\end{vmatrix}$

$=\hat{\text{i}}(\text{a}_2\text{b}_3-\text{a}_3\text{b}_2)-\hat{\text{j}}(\text{a}_1\text{b}_3-\text{a}_3\text{b}_1)+\hat{\text{k}}(\text{a}_1\text{b}_2-\text{a}_2\text{b}_1)$

$\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{b}}$

$=\big[\hat{\text{i}}(\text{a}_2\text{b}_3-\text{a}_3\text{b}_2)-\hat{\text{j}}(\text{a}_1\text{b}_3-\text{a}_3\text{b}_1)+\hat{\text{k}}(\text{a}_1\text{b}_2-\text{a}_2\text{b}_1)\big]\\.\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$

$=\text{a}_2\text{b}_1\text{b}_3-\text{a}_3\text{b}_1\text{b}_2-\text{a}_1\text{b}_2\text{b}_3+\text{a}_3\text{b}_1\text{b}_2+\text{a}_1\text{b}_2\text{b}_3-\text{a}_2\text{b}_1\text{b}_3$

$=\text{b}_1(\text{a}_2\text{b}_3-\text{a}_3\text{b}_2)-\text{b}_2(\text{a}_1\text{b}_3-\text{a}_3\text{b}_1)+\text{b}_3(\text{a}_1\text{b}_2-\text{a}_2\text{b}_1)$

$=\text{a}_2\text{b}_1\text{b}_3-\text{a}_3\text{b}_1\text{b}_2-\text{a}_1\text{b}_2\text{b}_3+\text{a}_3\text{b}_1\text{b}_2+\text{a}_1\text{b}_2\text{b}_3-\text{a}_2\text{b}_1\text{b}_3$

$=0$

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Question 972 Marks
If $\vec{\text{a}}=3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},$ then find $\big(\vec{\text{a}}\times\vec{\text{b}}\big)\vec{\text{a}}.$
Answer
Since $\vec{\text{a}}\times\vec{\text{b}}$ is a vector, $\big(\vec{\text{a}}\times\vec{\text{b}}\big)\vec{\text{a}}$ without any dot or cross product in between is meaningless.
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Question 982 Marks
If the position vector of a point (-4, -3) be $\vec{\text{a}}$, find $\big|\vec{\text{a}}\big|$.
Answer
Given a point (-4, -3) such that its position vector $\vec{\text{a}}$ is given by
$\vec{\text{a}}=-4\hat{\text{i}}-3\hat{\text{j}}$
Then,
$\big|\vec{\text{a}}\big|=\sqrt{(-4)^2+(-3)^2}$
$=\sqrt{16+9}$
$=\sqrt{25}$
$=5$
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Question 992 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are the position vectors of the vertices of a triangle, then write the position vector of its centroid.
Answer
Let ABC be a triangle and D, E and F are the midpoints of the sides BC, CA and AB respectively.
Also, Let $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are the position vectors  of A, B, C respectively. Then the position vectors of D, E, F are $\Big(\frac{\vec{\text{b}}+\vec{\text{c}}}2\Big),\Big(\frac{\vec{\text{c}}+\vec{\text{a}}}2\Big),\Big(\frac{\vec{\text{a}}+\vec{\text{b}}}2\Big)$ respectively.
The position vector of a point divides AD in the ratio of 2; is $\frac{1.\vec{\text{a}}+2\frac{\vec{\text{b}}+\vec{\text{c}}}{2}}{2}=\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3$
Similarly, Position vectors of the points divides BE, CF in the ratio of 2 : 1 are equal to $\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3$.
Thus, the point dividing AD in the ratio 2 : 1 also divides BE, CF in the same ratio.
Hence, the medians of a triangle are concurrent and the position vector of the centroid is $\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3$.
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Question 1002 Marks
Find the projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ if $\vec{\text{a}}.\vec{\text{b}}=8$ and $\vec{\text{b}}=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}.$
Answer
We 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
$\frac{​\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{b}}|}$
$=\frac{8}{\sqrt{4+36+9}}$
$=\frac{8}{7}$
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