MHT CET · Maths · Three Dimensional Geometry
If the foot of the perpendicular drawn from the origin to a plane is \(M(-1,-2,2)\), then the vector equation of the plane is
- A \(\bar{r} \cdot(-\hat{i}-2 \hat{j}+2 \widehat{k})=9\)
- B \(\bar{r} \cdot(\widehat{i}+2 \widehat{j}+2 \widehat{k})=9\)
- C \(\bar{r} \cdot(-\hat{i}-2 \hat{j}-2 \widehat{k})=9\)
- D \(\bar{r} \cdot(\hat{i}+2 \hat{j}-2 \widehat{k})=9\)
Answer & Solution
Correct Answer
(A) \(\bar{r} \cdot(-\hat{i}-2 \hat{j}+2 \widehat{k})=9\)
Step-by-step Solution
Detailed explanation
D.r's of normal to the plane \(<-1,-2,2>\)
D.c's of Normal to the plane \(<-\frac{1}{3},-\frac{2}{3}, \frac{2}{3}>\)
length of perpendicular \(=\sqrt{1^2+2^2+(-2)^2}=3\)
Hence, Equation of plane \(\vec{r} \cdot\left(-\frac{1}{3} \hat{i}-\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}\right)=3\)
\(\Rightarrow \vec{r} \cdot(-\hat{i}-2 \hat{j}+2 \widehat{k})=9\)
D.c's of Normal to the plane \(<-\frac{1}{3},-\frac{2}{3}, \frac{2}{3}>\)
length of perpendicular \(=\sqrt{1^2+2^2+(-2)^2}=3\)
Hence, Equation of plane \(\vec{r} \cdot\left(-\frac{1}{3} \hat{i}-\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}\right)=3\)
\(\Rightarrow \vec{r} \cdot(-\hat{i}-2 \hat{j}+2 \widehat{k})=9\)
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