A vector n of magnitude 8 units is inclined t | vedclass

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(B) Let $\gamma$ be the angle made by $n$ with the $z$-axis.The direction cosines of $n$ are $l = \cos 45^\circ = \frac{1}{\sqrt{2}}$,$m = \cos 60^\ci

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$r \cdot (\sqrt{2}i + j + k) = 2$

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(B) Let $\gamma$ be the angle made by $n$ with the $z$-axis.The direction cosines of $n$ are $l = \cos 45^\circ = \frac{1}{\sqrt{2}}$,$m = \cos 60^\circ = \frac{1}{2}$,and $n = \cos \gamma$.Since $l^2 + m^2 + n^2 = 1$,we have $(\frac{1}{\sqrt{2}})^2 + (\frac{1}{2})^2 + n^2 = 1$.$\frac{1}{2} + \frac{1}{4} + n^2 = 1 \Rightarrow n^2 = 1 - \frac{3}{4} = \frac{1}{4}$.Since $\gamma$ is acute,$n = \cos \gamma = \frac{1}{2}$.Given $|n| = 8$,the vector $n = |n|(li + mj + nk) = 8(\frac{1}{\sqrt{2}}i + \frac{1}{2}j + \frac{1}{2}k) = 4\sqrt{2}i + 4j + 4k$.The plane passes through $a = \sqrt{2}i - j + k$ and is normal to $n$. The equation is $r \cdot n = a \cdot n$.$r \cdot (4\sqrt{2}i + 4j + 4k) = (\sqrt{2}i - j + k) \cdot (4\sqrt{2}i + 4j + 4k)$.$r \cdot (4\sqrt{2}i + 4j + 4k) = (\sqrt{2})(4\sqrt{2}) + (-1)(4) + (1)(4) = 8 - 4 + 4 = 8$.Dividing by $4$,we get $r \cdot (\sqrt{2}i + j + k) = 2$.

$A$ vector $n$ of magnitude $8$ units is inclined to $x$-axis at $45^\circ$,$y$-axis at $60^\circ$ and an acute angle with $z$-axis. If a plane passes through a point $(\sqrt{2}, -1, 1)$ and is normal to $n$,then its equation in vector form is

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