If the position vectors of $P$ and $Q$ are $¯ i + 2 ¯ j - 7 ¯ ¯ ¯ k$ and $5 ¯ i - 3 ¯ j + 4 ¯ ¯ ¯ k$ respectively then the cosine of the angle between $¯ ¯¯¯¯¯¯ ¯ P Q$ and $z$ -axis is

\frac{4}{\sqrt{162}}

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\frac{11}{\sqrt{162}}

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\frac{5}{\sqrt{162}}

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\frac{- 5}{\sqrt{162}}

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Solution

The correct option is A $\frac{11}{\sqrt{162}}$
$\to P Q = \to Q - \to P$
$= 5 i - 3 j + 4 k - (i + 2 j - 7 k)$
$= 4 i - 5 j + 11 k$
Now
$Z_{c o m p o n e n t} = \to P Q . k$
$cos γ = \frac{\to P Q . k}{| P Q |}$
$= \frac{(3 i - 5 j + 11 k) . k}{\sqrt{4^{2} + 5^{2} + 11^{2}}}$
$= \frac{11}{\sqrt{162}}$

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Similar questions

\hat{i} + 3 \hat{j} - 7 \hat{k} and 5 \hat{i} - 2 \hat{j} + 4 \hat{k}

\vec{P Q}

\frac{5}{\sqrt{162}}

\frac{4}{\sqrt{162}}

- \frac{5}{\sqrt{162}}

\frac{11}{\sqrt{162}}

P = 116^{2} - 16^{2}, Q = 124^{2} - 24^{2}

R = 120

\sqrt{\frac{P}{100} + \frac{Q}{100} + R} =

5 \sqrt{2} - \sqrt{98} + \sqrt{162}

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