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The equation of the plane through $(-1,1,2)$, whose normal makes equal acute angles with coordinate axes is (A) $\bar{r} \cdot(\hat{i}+\hat{j}+\hat{k})=2$ (B) $\bar{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$ (C) $\bar{r} \cdot(3 \hat{i}-3 \hat{j}+3 \hat{k})=2$ (D) $\bar{r} \cdot(\hat{i}-\hat{j}+\hat{k})=3$
The equation of the plane through $(-1,1,2)$, whose normal makes equal acute angles with coordinate axes is
(A) $\bar{r} \cdot(\hat{i}+\hat{j}+\hat{k})=2$
(B) $\bar{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$
(C) $\bar{r} \cdot(3 \hat{i}-3 \hat{j}+3 \hat{k})=2$
(D) $\bar{r} \cdot(\hat{i}-\hat{j}+\hat{k})=3$
maths | Maths
The equation of the plane through $(-1,1,2)$, whose normal makes equal acute angles with coordinate axes is
(A) $\bar{r} \cdot(\hat{i}+\hat{j}+\hat{k})=2$
(B) $\bar{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$
(C) $\bar{r} \cdot(3 \hat{i}-3 \hat{j}+3 \hat{k})=2$
(D) $\bar{r} \cdot(\hat{i}-\hat{j}+\hat{k})=3$

Correct option is

(A) $\bar{r} \cdot(\hat{i}+\hat{j}+\hat{k})=2$

Equation plane passing through

$\mathrm{A}(\overrightarrow{\mathrm{a}})$ and $\perp$ to $\overrightarrow{\mathrm{n}}$ is

$\overrightarrow{\mathrm{r}} \cdot \hat{\mathrm{n}}=\overrightarrow{\mathrm{a}} \cdot \hat{\mathrm{n}}$

Here $\vec{a}=-\hat{i}+\hat{j}+2 \hat{k}, \vec{n}=\hat{i}+\hat{j}+\hat{k}$

$\therefore \overline{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=2$

 

i

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