If a- is perpendicularto b- and r- is a non-zero vector such that pr-r-b-a-c- - then r-

The correct option is A c⃗p (b⃗.c⃗)a⃗p^2 Taking dot product of b on both sides ⇒ p (r.b)+(r.b)(a.b)=c.b r.b=b.cp ∴ , r=cp (a.c).b ...

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Byju's Answer

Standard XII

Mathematics

Combination

If a⃗ is pe...

Question

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

\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 b) \to c}{p^{2}}

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

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Solution

The correct option is A $\frac{\to c}{p} - \frac{(\to b . \to c) \to a}{p^{2}}$
Taking dot product of $b$ on both sides
$\Rightarrow p (r . b) + (r . b) (a . b) = c . b$
$r . b = \frac{b . c}{p}$
$∴$ , $r = \frac{c}{p} - \frac{(a . c) . b}{p^{2}}$

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Similar questions

Q. 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

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\to r

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Q. If

\to a

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p \to r + (\to r . \to b) \to a = \to c

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\to a

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is a non – zero vector such that

p \to r + (\to r . \to b) \to a = \to c

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\to a, \to b, \to c

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Δ A B C

. If

\to r

is position vector of a point

P

such that

(| \to b - \to c | + | \to c - \to a | + | \to a - \to b |) \to r = | \to b - \to c | \to a + | \to c - \to a | \to b + ∣ ∣ \to a - \to b ∣ ∣ \to c

then the point

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Q. If

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\to p = \frac{\to b \times \to c}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}, \to q = \frac{\to c \times \to a}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}, \to r = \frac{\to a \times \to b}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}

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If →a is perpendicularto →b and →r is a non-zero vector such that p→r+(→r.→b)→a=→c , then →r=

The correct option is A →cp−(→b.→c)→ap2Taking dot product of b on both sides⇒p(r.b)+(r.b)(a.b)=c.br.b=b.cp∴, r=cp−(a.c).bp2

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

The correct option is A $\frac{\to c}{p} - \frac{(\to b . \to c) \to a}{p^{2}}$
Taking dot product of $b$ on both sides
$\Rightarrow p (r . b) + (r . b) (a . b) = c . b$
$r . b = \frac{b . c}{p}$
$∴$ , $r = \frac{c}{p} - \frac{(a . c) . b}{p^{2}}$