If the position vectors of P and Q are hat{i} | vedclass

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(B) Given the position vectors $\overrightarrow{OP} = \hat{i}+2 \hat{j}-7 \hat{k}$ and $\overrightarrow{OQ} = 5 \hat{i}-3 \hat{j}+4 \hat{k}$.First, we

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

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(B) Given the position vectors $\overrightarrow{OP} = \hat{i}+2 \hat{j}-7 \hat{k}$ and $\overrightarrow{OQ} = 5 \hat{i}-3 \hat{j}+4 \hat{k}$.First, we find the vector $\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}$.$\overrightarrow{PQ} = (5 \hat{i}-3 \hat{j}+4 \hat{k}) - (\hat{i}+2 \hat{j}-7 \hat{k}) = 4 \hat{i}-5 \hat{j}+11 \hat{k}$.The direction vector of the $z$-axis is $\hat{k} = 0 \hat{i} + 0 \hat{j} + 1 \hat{k}$.Let $	heta$ be the angle between $\overrightarrow{PQ}$ and the $z$-axis. The cosine of the angle is given by $\cos 	heta = \frac{\overrightarrow{PQ} \cdot \hat{k}}{|\overrightarrow{PQ}| |\hat{k}|}$.Calculating the dot product: $\overrightarrow{PQ} \cdot \hat{k} = (4)(0) + (-5)(0) + (11)(1) = 11$.Calculating the magnitude of $\overrightarrow{PQ}$: $|\overrightarrow{PQ}| = \sqrt{4^2 + (-5)^2 + 11^2} = \sqrt{16 + 25 + 121} = \sqrt{162}$.Since $|\hat{k}| = 1$, we have $\cos 	heta = \frac{11}{\sqrt{162} 	imes 1} = \frac{11}{\sqrt{162}}$.

If the position vectors of $P$ and $Q$ are $\hat{i}+2 \hat{j}-7 \hat{k}$ and $5 \hat{i}-3 \hat{j}+4 \hat{k}$ respectively, then the cosine of the angle between $\overrightarrow{PQ}$ and $z$-axis is

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