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Vector Algebra Part 2

Section: Mathematics

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Question : 53 of 100

Marks: +1, -0

Let

\overset{\rightarrow}{a}

be a vector which is perpendicular to the vector

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

\overset{\rightarrow}{a} \times (2 \hat{i} + \hat{k}) = 2 \hat{i} - 13 \hat{j} - 4 \hat{k}

, then the projection of the vector

\overset{\rightarrow}{a}

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

[28-Jun-2022-Shift-2]

$\frac{1}{3}$
1
$\frac{5}{3}$
$\frac{7}{3}$

Solution:

Let

\overset{\rightarrow}{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}

and

\overset{\rightarrow}{a} \cdot \left(3 \hat{i} - \frac{1}{2} \hat{j} + 2 \hat{k}\right) = 0 \Rightarrow 3 a_1 + \frac{a_2}{2} + 2 a_3 = 0 \ldots

. (i)

and

\overset{\rightarrow}{a} \times (2 \hat{i} + \hat{k}) = 2 \hat{i} - 13 \hat{j} - 4 \hat{k}

\Rightarrow a_2 \hat{i} + (2 a_3 - a_1) \hat{j} - 2 a_2 \hat{k} = 2 \hat{i} - 13 \hat{j} - 4 \hat{k}

\therefore a_2 = 2 \ldots

. (ii)

and

a_1 - 2 a_3 = 13 \ldots

. (iii)

From eq. (i) and (iii) :

a_1 = 3

and

a_3 = -5

\therefore \overset{\rightarrow}{a} = 3 \hat{i} + 2 \hat{j} - 5 \hat{k}

\therefore

projection of

\overset{\rightarrow}{a}

on

2 \hat{i} + 2 \hat{j} + \hat{k} = \frac{6+4-5}{3} = \frac{5}{3}

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