Let a -2k- b - b 1- b 2-2k- and c -5 -2k- be three vectors such that the projection vector of b on a is a- Ifa - b is perpendicular to c- then - b - is equal to -A- -32B- 4 - 6D- -22

Given, Projection of b on a is a. ∴b . a|a| = |a| ⇒b_1 + b_2 + 2√(1 + 1 + 2) = √(1 + 1 +2) ⇒ b_1 + b_2 = 2 nbsp;...(1) nbsp; nbsp; nbsp; nb ...

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Byju's Answer

Standard XII

Mathematics

Applications of Dot Product

Let a =î+ĵ+√2...

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 $\to a$ . If $\to a + \to b$ is perpendicular to $\to c$ , then $| \to b |$ is equal to :

6

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4

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\sqrt{22}

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\sqrt{32}

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Solution

The correct option is A $6$
Given, Projection 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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Similar questions

Q. Let

\to a =^i +^j + \sqrt{2}^k, \to b = b_{1}^i + b_{2}^j + \sqrt{2}^k

and

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Q. Let

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\to a = \to i + \to j + \sqrt{1}^k, \to b = b_{1}^i + b_{1}^j + \sqrt{2}^k

and

\to c = 5^i +^j + \sqrt{2}^k

be three vectors such that the projection vector of

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Let →a=^i+^j+√2^k,→b=b1^i+b2^j+√2^k and →c=5^i+^j+√2^k be three vectors such that the projection vector of →b on →a is →a. If →a+→b is perpendicular to →c, then |→b| is equal to :

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 $\to a$ . If $\to a + \to b$ is perpendicular to $\to c$ , then $| \to b |$ is equal to :