Let bar{n} be a vector of magnitude 3sqrt{3} | vedclass

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(D) Let the vector $\bar{n}$ be $a\hat{i} + a\hat{j} + a\hat{k}$ since it makes equal angles with the coordinate axes. Given magnitude $|\bar{n}| = 3\

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$\bar{r} \cdot (3\hat{i} + 3\hat{j} + 3\hat{k}) = 6$

Vedclass · Answer

(D) Let the vector $\bar{n}$ be $a\hat{i} + a\hat{j} + a\hat{k}$ since it makes equal angles with the coordinate axes. Given magnitude $|\bar{n}| = 3\sqrt{3}$,so $\sqrt{a^2 + a^2 + a^2} = 3\sqrt{3} \Rightarrow \sqrt{3a^2} = 3\sqrt{3} \Rightarrow a\sqrt{3} = 3\sqrt{3} \Rightarrow a = 3$. Thus,$\bar{n} = 3\hat{i} + 3\hat{j} + 3\hat{k}$. The equation of a plane passing through point $\vec{a} = (1, -1, 2)$ and normal to $\bar{n}$ is given by $\vec{r} \cdot \bar{n} = \vec{a} \cdot \bar{n}$. $\vec{r} \cdot (3\hat{i} + 3\hat{j} + 3\hat{k}) = (1\hat{i} - 1\hat{j} + 2\hat{k}) \cdot (3\hat{i} + 3\hat{j} + 3\hat{k})$. $\vec{r} \cdot (3\hat{i} + 3\hat{j} + 3\hat{k}) = (1)(3) + (-1)(3) + (2)(3) = 3 - 3 + 6 = 6$. Therefore,the equation is $\bar{r} \cdot (3\hat{i} + 3\hat{j} + 3\hat{k}) = 6$.

Let $\bar{n}$ be a vector of magnitude $3\sqrt{3}$ such that it makes equal acute angles with the coordinate axes. Then the vector equation of a plane passing through $(1, -1, 2)$ and normal to $\bar{n}$ is:

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