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Let a = î + ĵ...

Question

Let $\to a =^i +^j + \sqrt{2}^k, \to b = b_{1}^i + b_{2}^j + \sqrt{2}^k$ and $\to c = 5^i +^j + \sqrt{2}^k$ be three vectors such that the projection vector of $\to b$ on $\to a$ is $\to a$ and $(\to a + \to b)$ is perpendicular to $\to c$ , then $| \to b |$ is equal to

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Solution

Given, Projection vector of $\to b$ on $\to a$ is $\to a$ .
$∴ \frac{\to b . \to a}{| \to a |} = | \to a |$

$\Rightarrow \frac{b_{1} + b_{2} + 2}{\sqrt{1 + 1 + 2}} = \sqrt{1 + 1 + 2}$

$\Rightarrow b_{1} + b_{2} = 2 . . . (1)$

Since, $\to a + \to b$ is perpendicular to $\to c$
So, $(\to a + \to b) \cdot \to c = 0$
$\Rightarrow \to a \cdot \to c + \to b \cdot \to c = 0$
$\Rightarrow (5 + 1 + 2) + (5 b_{1} + b_{2} + 2) = 0$
$\Rightarrow 5 b_{1} + b_{2} = - 10 . . . (2)$

From Equations $(1)$ and $(2)$ ,
$b_{1} = - 3, b_{2} = 5$
$∴ \to b = - 3^i + 5^j + \sqrt{2}^k$
Hence, $| \to b | = \sqrt{9 + 25 + 2}$
$\Rightarrow | \to b | = 6$

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