If a- is perpendicular to b- and r- is a non - zero vector such that p r-r-b-a-c- then r- is equal to

We have, pr⃗+(r⃗.b⃗)a⃗=c⃗ ⇒ p(r⃗.b⃗)+(a⃗.b⃗)(a⃗.b⃗)=c⃗.b⃗ ⇒ p(r⃗.b⃗)=c⃗.b⃗ [∵a⃗⊥b⃗ , a⃗.b⃗=0] ⇒r⃗.b⃗=c⃗.b⃗/p Substituting the value of r⃗.b⃗ in Eq. (i), we get ...

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Byju's Answer

Standard XI

Mathematics

Applications of Dot Product

If a⃗ is perp...

Question

If $\to a$ is perpendicular to $\to b$ and $\to r$ is a non – zero vector such that $p \to r + (\to r . \to b) \to a = \to c$ , then $\to r$ is equal to

\frac{\to c}{p} - \frac{(\to b . \to c) \to a}{p^{2}}

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\frac{\to a}{p} - \frac{(\to c . \to a) \to b}{p^{2}}

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\frac{\to b}{p} - \frac{(\to a . \to c) \to c}{p^{2}}

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\frac{\to c}{p^{2}} - \frac{(\to b . \to c) \to a}{p}

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Solution

The correct option is A $\frac{\to c}{p} - \frac{(\to b . \to c) \to a}{p^{2}}$
We have, $p \to r + (\to r . \to b) \to a = \to c$
$\Rightarrow p (\to r . \to b) + (\to a . \to b) (\to a . \to b) = \to c . \to b \Rightarrow p (\to r . \to b) = \to c . \to b [∵ \to a ⊥ \to b, \to a . \to b = 0]$
$\Rightarrow \to r . \to b = \frac{\to c . \to b}{p}$
Substituting the value of $\to r . \to b$ in Eq. (i), we get
$p \to r + (\frac{\to c . \to b}{p}) \to a = \to c \Rightarrow \to r = \frac{\to c}{p} - \frac{\to c . \to b}{p^{2}} \to a$

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If →a is perpendicular to →b and →r is a non – zero vector such that p→r+(→r.→b)→a=→c, then →r is equal to

If $\to a$ is perpendicular to $\to b$ and $\to r$ is a non – zero vector such that $p \to r + (\to r . \to b) \to a = \to c$ , then $\to r$ is equal to