A vector $\vec{n}$ is inclined to $X$-axis at $45^{\circ}$, $\mathrm{Y}$-axis at $60^{\circ}$ and at an…

A vector $\vec{n}$ is inclined to $X$-axis at $45^{\circ}$, $\mathrm{Y}$-axis at $60^{\circ}$ and at an acute angle to $\mathrm{Z}$-axis. If $\overrightarrow{\mathrm{n}}$ is normal to a plane passing through the point $(-\sqrt{2}, 1,1)$, then equation of the plane is
  1. $\sqrt{2} x+y+z=0$
  2. $x+\sqrt{2} y+z=1$
  3. $-\sqrt{2} x+y+2 z=5$
  4. $x+y+\sqrt{2} z=1$

Solution

Let $\vec{n}$ be inclined at angles $\alpha, \beta, \gamma$ to $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ axes respectively. $\begin{array}{ll} & \alpha=45^{\circ}, \beta=60^{\circ}, \gamma=? \\ & \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \\ \therefore \quad & \frac{1}{2}+\frac{1}{4}+\cos ^2 \gamma=1 \\ \therefore \quad & \cos ^2 \gamma=\frac{1}{4} \\ \therefore \quad & \gamma=60^{\circ} \\ & \overrightarrow{\mathrm{n}}=\cos \alpha \hat{\mathrm{i}}+\cos \beta \hat{\mathrm{j}}+\cos \gamma \hat{\mathrm{k}} \\ \therefore \quad & \overrightarrow{\mathrm{n}}=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}+\frac{1}{2} \hat{\mathrm{k}} \end{array}$ $\therefore \quad$ Equation of the required plane is $\frac{1}{\sqrt{2}}(x+\sqrt{2})+\frac{1}{2}(y-1)+\frac{1}{2}(z-1)=0$ i.e., $\sqrt{2} x+y+z=0$

Asked in: MHT CET 2023 (09 May Shift 1)

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