If $3 p + 2 q = i + j + k$ and $3 p - 2 q = i - j - k$ , then the angle between $p$ and $q$ is

\frac{π}{6}

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\frac{π}{4}

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\frac{π}{3}

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\frac{π}{2}

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π

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Solution

The correct option is D $\frac{π}{2}$
Given, $3 p + 2 q = i + j + k$ ..... (i)
and $3 p - 2 q = i - j - k$ .... (ii)
On adding Eqs. (i) and (ii), we get
$6 p = 2 i$
Therefore, $p = \frac{i}{3}$
On subtracting Eq. (ii) from Eq. (i), we get
$4 q - 2 (j + k) \Rightarrow q = \frac{1}{2} (j + k)$
Let $θ$ be the angle between $p$ and $q$ , then
$p \cdot q = | p | | q | cos θ$
$\Rightarrow cos θ = \frac{p \cdot q}{| p | | q |}$
$\Rightarrow cos θ = \frac{1}{6} \frac{i \cdot (j + k)}{| p | | q |} = 0 = cos 90^{\circ}$
Thus $θ = 90^{\circ} = \frac{π}{2}$

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If 3p+2q=i+j+k and 3p−2q=i−j−k, then the angle between p and q is

If $3 p + 2 q = i + j + k$ and $3 p - 2 q = i - j - k$ , then the angle between $p$ and $q$ is