MHT CET · Maths · Three Dimensional Geometry
The equation of a plane containing the point \((1,-1,1)\) and parallel to the plane \(2 x+3 y-4 z=17\) is
- A \(\overline{\mathrm{r}} \cdot(2 \hat{\imath}-3 \hat{\jmath}-4 \hat{\mathrm{k}})=-1\)
- B \(\overline{\mathrm{r}} \cdot(\hat{\imath}-\hat{\mathrm{j}}+\hat{\mathrm{k}})=3\)
- C \(\overline{\mathrm{r}} \cdot(2 \hat{\imath}+3 \hat{\jmath}-4 \hat{\mathrm{k}})=-5\)
- D \(\overline{\mathrm{r}} \cdot(2 \hat{\imath}+3 \hat{\jmath}-4 \hat{\mathrm{k}})=5\)
Answer & Solution
Correct Answer
(C) \(\overline{\mathrm{r}} \cdot(2 \hat{\imath}+3 \hat{\jmath}-4 \hat{\mathrm{k}})=-5\)
Step-by-step Solution
Detailed explanation
Here \(\bar{a}=\hat{i}-\hat{j}+\hat{k}\)
\(\& \quad \bar{n}=2 \hat{i}+3 \hat{j}-4 \hat{k}\)
Now \(\bar{a} \cdot \bar{n} \quad=2-3-4=-5\)
Vector equation of plane passing through \(\mathrm{A}(\bar{a})\) is
\(
\begin{aligned}
& \overline{\mathrm{r}} \cdot \overline{\mathrm{n}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{n}} \\
\Rightarrow & \overline{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})=-5
\end{aligned}
\)
\(\& \quad \bar{n}=2 \hat{i}+3 \hat{j}-4 \hat{k}\)
Now \(\bar{a} \cdot \bar{n} \quad=2-3-4=-5\)
Vector equation of plane passing through \(\mathrm{A}(\bar{a})\) is
\(
\begin{aligned}
& \overline{\mathrm{r}} \cdot \overline{\mathrm{n}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{n}} \\
\Rightarrow & \overline{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})=-5
\end{aligned}
\)
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