MHT CET · Maths · Three Dimensional Geometry
The equation of the plane through whose normal makes equal acute angle with co-ordinate axes is
- A
- B
- C
- D
Answer & Solution
Correct Answer
(A)
Step-by-step Solution
Detailed explanation
Equation plane passing through
\(A(\vec{a})\) and \(\perp\) to \(\vec{n}\) is
\(\vec{r} \cdot \vec{n}=\vec{a} \cdot \vec{n}\)
Here \(\vec{a}=-\hat{i}+\hat{j}+2 \hat{k}, \quad \vec{n}=\hat{i}+\hat{j}+\hat{k}\)
\(\therefore \vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=(-\hat{i}+\hat{j}+2 \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})=2\)
\(A(\vec{a})\) and \(\perp\) to \(\vec{n}\) is
\(\vec{r} \cdot \vec{n}=\vec{a} \cdot \vec{n}\)
Here \(\vec{a}=-\hat{i}+\hat{j}+2 \hat{k}, \quad \vec{n}=\hat{i}+\hat{j}+\hat{k}\)
\(\therefore \vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=(-\hat{i}+\hat{j}+2 \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})=2\)
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