If p- s- are not perpendicular to each other and r-p-q-p- r-s-0- then r-

The correct option is B q⃗ ( q⃗.s⃗/p⃗.s⃗ )p⃗ Let r,p,q and s be vectors.It is given that r× p=q× p r× p (q× p)=0 (r q)× p=0 Hen ...

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Standard XII

Mathematics

Definition of Sets

If p⃗ s...

Question

If $\to p$ & $\to s$ are not perpendicular to each other and $\to r \times \to p = \to q \times \to p$ & $\to r . \to s = 0,$ then $\to r =$

\to p . \to s

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\to q + (\frac{\to q . \to s}{\to p . \to s}) \to p

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\to q - (\frac{\to q . \to s}{\to p . \to s}) \to p

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\to q + μ \to p

for all scalars

μ

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Solution

The correct option is B $\to q - (\frac{\to q . \to s}{\to p . \to s}) \to p$
Let $r, p, q$ and $s$ be vectors.
It is given that
$r \times p = q \times p$
$r \times p - (q \times p) = 0$
$(r - q) \times p = 0$
Hence $r - q$ is a vector parallel to vector $p$ .
Hence
$r = q + μ p$ ...(i)
It is given that $r . s = 0$ .
Hence $r$ vector is perpendicular to $s$ vector.
Hence $(q + μ p) . s = 0$ ...(from i)
$q . s + μ (p . s) = 0$
$μ (p . s) = - q . s$
$μ = - \frac{q . s}{p . s}$
Hence $r = q - \frac{q . s}{p . s} p$

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If →p & →s are not perpendicular to each other and →r×→p=→q×→p & →r.→s=0, then →r=

If $\to p$ & $\to s$ are not perpendicular to each other and $\to r \times \to p = \to q \times \to p$ & $\to r . \to s = 0,$ then $\to r =$