Case study problem vector 1 chapter 10 class 12

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Case study Chapter 10 (Vectors)

Solar panels have to be installed carefully so that the tilt of the roof, and the direction to the sun , produce the largest possible electrical power in the solar panals.(Case study problem vector 1)

A surveyor uses his instrument to determine the co-ordinates of the four corners of a roof where solar panels are to

be mounted. In the picture, suppose the points are labelled counter clockwise from the roof corner nearest to the

camera in units of meters $P_1(6, 8, 4),P_2(21, 8, 4),P_3(21, 16, 10)$ and $P_4(6, 16, 10)$ .

Based on the above information answer the following question.

(i) Find the components to the two edge vectors defined by $\vec{A} = \vec{OP_2}-\vec{OP_1}$ and $\vec{B} = \vec{OP_4}-\vec{OP_1}$ ?

(ii) (a) Find the magnitudes of the vectors $\vec{A}$ and $\vec{B}$ .

(b) Find the components to the vectors $\vec{N}$ , perpendicular to $\vec{A}$ and $\vec{B}$ and the surface of the roof.

Solution: Given points are $P_1(6, 8, 4),P_2(21, 8, 4),P_3(21, 16, 10)$ and $P_4(6, 16, 10)$ .

(i) We have,

$\vec{A} = \vec{OP_2}-\vec{OP_1}$

$= (21\vec{i}+8\vec{j}+4\vec{k})- (6\vec{i}+8\vec{j}+4\vec{k})$

$\Rightarrow \vec{A} = (15\vec{i}+0\vec{j}+0\vec{k})$

Components of vector A are $15, 0 , 0$

and $\vec{B} = \vec{OP_4}-\vec{OP_1}$

$= (6\vec{i}+16\vec{j}+10\vec{k})- (6\vec{i}+8\vec{j}+4\vec{k})$

$\Rightarrow \vec{B} = (0\vec{i}+8\vec{j}+6\vec{k})$

Components of vector B are $0, 8 , 6$ .

(ii) (a) We have

$|A| = \sqrt{(15)^2+(0)^2+(0)^2} = 15$ units

$|B| = \sqrt{(0)^2+(8)^2+(6)^2} = 10$ units

(b) We have,

$\vec{N} = \vec{A}\times \vec{B}$

$= \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\15 & 0 & 0 \\0 & 8 & 6\end{vmatrix}$

$= \vec{i}(0-0)-\vec{j}(90-0)\vec{k}(120-0)$

$= 0\vec{i}-90\vec{j}+120\vec{k}$

Its components are $0 ,-90, 120$

2. Read the following and answer the question:

A class XII student appearing for a competitive examination was asked to attempt the following question.

Let $\vec{a},\vec{b}$ and $\vec{c}$ be three non zero vectors.

(i) If $\vec{a}$ and $\vec{b}$ are such that $|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|$

(a) $a\perp b$ (b) $\vec{a}||\vec{b}$

(ii) If $\vec{a}=\hat{i}-2\hat{j},\vec{b}=2\vec{i}+\vec{j}+3\vec{k}$ then evaluate $(2\vec{a}+\vec{b [(\vec{a}+\vec{b})\times (\vec{a}-2\vec{b})]$

(a) 0 (b) 4

(iii) If $\vec{a}$ and $\vec{b}$ are unit vectors and $\theta$ be the angle between them then $|\vec{a}-\vec{b}|$ is

(a) $\sin\frac{\theta}{2}$ (b) $2\sin\frac{\theta}{2}$

(iv) Let $\vec{a},\vec{b}$ and $\vec{c}$ be unit vectors such that $\vec{a}.\vec{b}=\vec{a}.\vec{c}=0$ and angle

between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{6}$ then $\vec{a}=$

(a) $2(\vec{b}\times \vec{c})$ (b) $-2(\vec{b}\times \vec{c})$

(v) The area of parallelogram formed by $\vec{a} = \hat{i}-2\hat{j}$ and $\vec{b}=2\hat{i}+\hat{j}+3\hat{k}$ as

diagonals is

(a) 70 (b) 35

Solution:(i) Answer (a)

Given, $|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|$

$\Rightarrow |\vec{a}+\vec{b}|^2=|\vec{a}-\vec{b}|^2$

$\Rightarrow |\vec{a}|^2+|\vec{b}|^2+2\vec{a}.\vec{b}=|\vec{a}|^2+|\vec{b}|^2-2\vec{a}.\vec{b}$

$\Rightarrow 4\vec{a}.\vec{b}=0$

$\Rightarrow \vec{a}.\vec{b}= 0$

$\Rightarrow |\vec{a}||\vec{b}|\cos\theta=0$

$\Rightarrow \cos\theta = 0$

$\Rightarrow \theta = \frac{\pi}{2}$

$\therefore \vec{a}\perp \vec{b} = 0$

(ii) Answer (a)

$\vec{a}=\hat{i}-2\hat{j},\vec{b}=2\vec{a}+\vec{j}+3\vec{k}$

$\therefore 2\vec{a}+\vec{b} = 2\hat{i}-4\hat{j}+2\vec{i}+\vec{j}+3\vec{k}$

$= 4\vec{i}+3\vec{j}+3\vec{k}$

$\vec{a}+\vec{b}=\hat{i}-2\hat{j}+2\vec{i}+\vec{j}+3\vec{k}$

$= 3\vec{i}-\vec{j}+3\vec{k}$

And
$\vec{a}-2\vec{b} = (\hat{i}-2\hat{j})-2(2\vec{i}+\vec{j}+3\vec{k})$

$= \hat{i}-2\hat{j}-4\vec{i}-2\vec{j}-6\vec{k}$

$= -3\hat{i}-4\hat{j}-6\hat{k}$

$\therefore (\vec{a}+\vec{b})\times (\vec{a}-2\vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\3 & -1 & 3\\ -3 & -4 & -6\end{vmatrix}$

$= \hat{i}(6+12)-\hat{j}(-18+9)+\hat{k}(-12-3)$

$= 18\hat{i}+9\hat{j}-15\hat{k}$

$\therefore (2\vec{a}+\vec{b}).[(\vec{a}+\vec{b})\times (\vec{a}-2\vec{b})] = (4\vec{i}+3\vec{j}+3\vec{k}).(18\hat{i}+9\hat{j}-15\hat{k})$

$= 4\times 18 - 3\times 9+3\times (-15)$

$= 72-27-45 = 72-72=0$

(iii) Answer(b)

Given, $|\vec{a}|=|\vec{b}| = 1$

$\therefore |\vec{a}-\vec{b}|^2 = |\vec{a}|^2+|\vec{b}|^2-2\vec{a}.\vec{b}$

$\Rightarrow |\vec{a}-\vec{b}|^2 = 1 + 1 - 2|\vec{a}||\vec{b}|\cos\theta$

$\Rightarrow |\vec{a}-\vec{b}|^2 = 2-2\cos\theta$

$\Rightarrow |\vec{a}-\vec{b}|^2 = 2(1-\cos\theta)$

$\Rightarrow |\vec{a}-\vec{b}|^2 = 2.2\sin^2\frac{\theta}{2}$

$\Rightarrow |\vec{a}-\vec{b}| = 2\sin\frac{\theta}{2}$

(iv) Answer (c)

We have $\vec{a}.\vec{b}=\vec{a}.\vec{c} = 0$

$\Rightarrow \vec{a}.\vec{b} = 0 \Rightarrow \vec{a}\perp\vec{b}$
And $\Rightarrow \vec{a}.\vec{c} = 0 \Rightarrow \vec{a}\perp\vec{c}$

Since angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{6}$

$\therefore \vec{a} =\pm2(\vec{b}\times \vec{c})$

(v) Answer (c)

The area of the parallelogram formed by $\vec{a}$ and $\vec{b}$ as diagonal $=\frac{1}{2}|\vec{a}\times\vec{b}|$

$\therefore A = \frac{1}{2}|\vec{a}\times\vec{b}| ---(i)$

Now, $\vec{a}\times\vec{b}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\1 & -2 & 0\\ 2 & 1 & 3\end{vmatrix}$

$= \hat{i}(-6-0)-\hat{j}(3-0)+\hat{k}(1+4)$

$= -6\hat{i}-3\hat{j}+5\hat{k}$

$|\vec{a}\times\vec{b}| = \sqrt{(-6)^2+(-3)^2+(5)^2}$

$= \sqrt{36+9+25}=\sqrt{70}$

Area $= \frac{1}{2}\sqrt{70}$ sq.units.

Some other case study question

Case study: A man is watching an aeroplane which is at the co-ordinate point A(4, -1, 3) assuming that the man is at O(0, 0, 0). At the same time he saw a bird at the coordinate point B(2, 0, 4).

Based on the above information answer the following:

(a) Find the position vector $\vec{AB}$