Case study problem vector 1 chapter 10 class 12

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}

(c) \vec{a}=\vec{b}   (d) None of these

(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

(c) 3                           (d) 2

(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}

(c) 2\cos\frac{\theta}{2} (d) \cos\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})

(c) \pm2(\vec{b}\times \vec{c})     (d) 2(\vec{b}\pm\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

(c) \frac{\sqrt{70}}{2}       (d) \sqrt{70}

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)


\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}


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

\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}

(b) Find the distance between aeroplane and bird

(c) Find the unit vector along \vec{AB} .

(d) Find the direction cosine of \vec{AB} .

(e) Find the angles which \vec{AB} makes with x, y and z axes.

Solution: For solution click here


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