AP EAMCET · Maths · Vector Algebra
The distance of a point \(\overrightarrow{\mathrm{a}}\) from the plane \(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{m}}=\mathrm{q}\) is given by \(\frac{|\vec{a} \cdot \vec{m}-q|}{|m|}\). If the distance of the point \(\hat{i}+2 \hat{j}+3 \hat{k}\) from the plane \(\vec{r} \cdot(2 \hat{i}+6 \hat{j}-9 \hat{k})=-1\) is \(p\) and the distance of the origin from this plane is \(\mathrm{q}\), then \(\mathrm{p}-\mathrm{q}=\)
- A \(6\)
- B \(5\)
- C \(2\)
- D \(1\)
Answer & Solution
Correct Answer
(D) \(1\)
Step-by-step Solution
Detailed explanation
Distance from point \(\vec{i}+2 \vec{j}+3 \vec{k}\) from the plane \(\begin{aligned} & \vec{r} \cdot(2 \vec{i}+b \vec{j}-9 \vec{k})=-1 \text { is qp } \\ & \Rightarrow \frac{|2+12-27+1|}{\sqrt{4+36+81}}=\frac{12}{11}=p \end{aligned}\) and distance from origin to the given plane…
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