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

Relation, r×p=q×p ⇒r×p q×p=0 ⇒ (r q) ×p=0 ⇒ (r q) and p are parallel ∴r q=λp ⇒r=λp+q …(1) Now, ⇒r·s=0 ⇒ (λp+q)·s=0 ⇒λ= q·sp·s …(2) nbsp; Putting λ from (2 ...

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Byju's Answer

Standard XII

Mathematics

Condition for Two Lines to be on the Same Plane

If p and s ar...

Question

If $\to p$ and $\to s$ are not perpendicular to each other and $\to r \times \to p = \to q \times \to p$ and $\to r \cdot \to s = 0,$ then $\to r$ is equal to

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

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\to q + μ \to p

for all

μ

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

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\to p \cdot \to s

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Solution

The correct option is C $\to q - (\frac{\to q \cdot \to s}{\to p \cdot \to s}) \to p$
relation,
$\to r \times \to p = \to q \times \to p$
$\Rightarrow \to r \times \to p - \to q \times \to p = 0$
$\Rightarrow (\to r - \to q) \times \to p = 0$
$\Rightarrow (\to r - \to q)$ and $\to p$ are parallel
$∴ \to r - \to q = λ \to p$
$\Rightarrow \to r = λ \to p + \to q \dots (1)$
Now, $\Rightarrow \to r \cdot \to s = 0$
$\Rightarrow (λ \to p + \to q) \cdot \to s = 0$
$\Rightarrow λ = - \frac{\to q \cdot \to s}{\to p \cdot \to s} \dots (2)$
Putting $λ$ from $(2)$ into $(1),$ we get
$\to r = - \frac{\to q \cdot \to s}{\to p \cdot \to s} \to p + \to q$
$∴ \to r = \to q - (\frac{\to q \cdot \to s}{\to p \cdot \to s}) \to p$

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If →p and →s are not perpendicular to each other and →r×→p=→q×→p and →r⋅→s=0, then →r is equal to

If $\to p$ and $\to s$ are not perpendicular to each other and $\to r \times \to p = \to q \times \to p$ and $\to r \cdot \to s = 0,$ then $\to r$ is equal to