If q- r- are unit vectors such that p-q-p-r- then

The correct options are A p⃗.q⃗=r⃗.q⃗ B maximum value of [p⃗ q⃗ r⃗]=1/2 C minimum value of [p⃗ q⃗ r⃗]=0 Take dot product with q⃗ on #16 ...

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Byju's Answer

Standard XII

Physics

Phase and the General Equation

If q⃗, r⃗ a...

Question

If $\to q, \to r$ are unit vectors such that $\to p = \to q \times \to p + \to r$ , then

\to p . \to q = \to r . \to q

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maximum value of

[\to p \to q \to r] = \frac{1}{2}

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minimum value of

[\to p \to q \to r] = 0

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minimum value of

[\to p \to q \to r] = - 1

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Solution

The correct options are
A $\to p . \to q = \to r . \to q$
B maximum value of $[\to p \to q \to r] = \frac{1}{2}$
C minimum value of $[\to p \to q \to r] = 0$
Take dot product with $\to q$ on both the sides
$\to p . \to q = \to r . \to q$
Hence, A is correct.
We have,
$\to p - \to q \times \to p = \to r$ - (i)
We have, $| \to r | = | \to q | = 1$
On squaring (i), we get
$p^{2} + p^{2} {sin}^{2} θ = 1$ ( $θ$ is the angle between the vectors $p$ and $q$ )
$p^{2} = \frac{1}{1 + {sin}^{2} θ}$
$\to p = \to q \times \to p + \to r$
Squaring $p^{2} = p^{2} {sin}^{2} θ + 2 [\begin{matrix} \to q \to p \to r \end{matrix}] + 1$
$2 [\begin{matrix} \to p \to q \to r \end{matrix}] = 1 - p^{2} {cos}^{2} θ = 1 - \frac{{cos}^{2} θ}{(1 + {sin}^{2} θ)}$
$2 [\begin{matrix} \to p \to q \to r \end{matrix}] = 1 - \frac{(1 - {sin}^{2} θ)}{(1 + {sin}^{2} θ)} = \frac{2 {sin}^{2} θ}{(1 + {sin}^{2} θ)}$
$2 [\begin{matrix} \to p \to q \to r \end{matrix}] = 2 (\begin{matrix} 1 - \frac{1}{1 + {sin}^{2} θ} \end{matrix})$
$[\begin{matrix} \to p \to q \to r \end{matrix}] = 1 - \frac{1}{(1 + {sin}^{2} θ)}$
When ${sin}^{2} θ = 0 \Rightarrow [\begin{matrix} \to p \to q \to r \end{matrix}] = 0$
When ${sin}^{2} θ = 1 \Rightarrow [\begin{matrix} \to p \to q \to r \end{matrix}] = \frac{1}{2}$ .

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If →q,→r are unit vectors such that →p=→q×→p+→r, then

If $\to q, \to r$ are unit vectors such that $\to p = \to q \times \to p + \to r$ , then