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KEAM 2015 Math Paper

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Question : 92 of 120

Marks: +1, -0

Let

\overline{OB} = \hat{i} + 2\hat{j} + 2\hat{k}

\overline{OA} = 4\hat{i} + 2\hat{j} + 2\hat{k}.

The distance of the point B from the straight line passing through A and parallel to the vector

2\hat{i} + 3\hat{j} + 6k

is

$\frac{7\sqrt{5}}{9}$
$\frac{5\sqrt{7}}{9}$
$\frac{3\sqrt{5}}{7}$
$\frac{9\sqrt{5}}{7}$
$\frac{9\sqrt{7}}{5}$

Solution:

Let the equation of line be

r = a + \lambda b_{2}

since, line passes through

A,

so

a = 4\hat{i} + 2\hat{j} + 2\hat{k}

and line is parallel to the vector

2\hat{i} + 3j + 6\hat{k}

so

b = 2\hat{i} + 3\hat{j} + 6\hat{k}

Hence, equation of the line is

r = 4\hat{i} + 2\hat{j} + 2\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})

Distance of point

B

from the line

= \left|\frac{\left| (a_2 - a_1) \times b \right|}{\left| b \right|}\right|

a_2 - a_1 = 3\hat{i} + 0\hat{j} + 0\hat{k}

(a_2 - a_1) \times b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & 0 \\ 2 & 3 & 6 \end{vmatrix} = -18\hat{j} + 9\hat{k}

|(a_2 - a_1) \times b| = \sqrt{(-18)^2 + (9)^2}

= \sqrt{405} = 9\sqrt{5}

|b| = \sqrt{4 + 9 + 36} = 7

Required distance

= \frac{9\sqrt{5}}{7}

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