MHT CET · Maths · Three Dimensional Geometry
The vector equation of plane containing the pint \((1,-1,2)\) and perpendicular to planes \(2 x+3 y-2 z=5\) and \(x+2 y-3 z=8\) is
- A \(\vec{r} \cdot(-5 \hat{i}+4 \hat{j}+\hat{k})=7\)
- B \(\vec{r} \cdot(-5 \hat{i}+4 \hat{j}-\hat{k})=-7\)
- C \(\vec{r} \cdot(-5 \hat{i}+4 \hat{j}+\hat{k})=-7\)
- D \(\vec{r} \cdot(-5 \hat{i}+4 \hat{j}-\widehat{k})=7\)
Answer & Solution
Correct Answer
(C) \(\vec{r} \cdot(-5 \hat{i}+4 \hat{j}+\hat{k})=-7\)
Step-by-step Solution
Detailed explanation
D.R's of the required plane can be obtained by
\(\frac{a}{3 \times(-3)-2 \times(-2)}=\frac{b}{1 \times(-2)-2 \times(-3)}=\) \(\frac{c}{2 \times 2-1 \times 3} \)
\( \Rightarrow \frac{a}{-5}=\frac{b}{4}=\frac{c}{1}\)
\(\Rightarrow\) the required equation is \(-5(x-1)+4(y+1)+1(z-2)=0\)
\(\Rightarrow-5 x+4 y+z=-7\)
\(\Rightarrow \vec{r} \cdot(-5 \hat{i}+4 \hat{j}+\hat{k})=-7\)
\(\frac{a}{3 \times(-3)-2 \times(-2)}=\frac{b}{1 \times(-2)-2 \times(-3)}=\) \(\frac{c}{2 \times 2-1 \times 3} \)
\( \Rightarrow \frac{a}{-5}=\frac{b}{4}=\frac{c}{1}\)
\(\Rightarrow\) the required equation is \(-5(x-1)+4(y+1)+1(z-2)=0\)
\(\Rightarrow-5 x+4 y+z=-7\)
\(\Rightarrow \vec{r} \cdot(-5 \hat{i}+4 \hat{j}+\hat{k})=-7\)
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