TS EAMCET · Maths · Three Dimensional Geometry
Let A be a point having position vector \(\bar{i}-3 \bar{j}\) and \(\bar{r}=(\bar{i}-3 \bar{j})+t(\bar{j}-2 \bar{k})\) be a line. If P is a point on this line and is at a minimum distance from the plane \(\bar{r} \bullet(2 \bar{i}+3 \bar{j}+5 \bar{k})=0\), then the equation of the plane through P and perpendicular to AP, is
- A \(\bar{r} \cdot(-\bar{j}+2 \bar{k})=8\)
- B \(\bar{r} \cdot(\bar{j}+\bar{k})=4\)
- C \(\bar{r} \cdot(\bar{i}+\bar{j}+\bar{k})=8\)
- D \(\bar{r} \cdot(\bar{i}-\bar{j})=12\)
Answer & Solution
Correct Answer
(A) \(\bar{r} \cdot(-\bar{j}+2 \bar{k})=8\)
Step-by-step Solution
Detailed explanation
\((\bar{i} + (-3+t)\bar{j} - 2t\bar{k}) \bullet (2 \bar{i}+3 \bar{j}+5 \bar{k}) = 0\) \(2 + 3(-3+t) + 5(-2t) = 0\) \(2 - 9 + 3t - 10t = 0 \Rightarrow -7 - 7t = 0 \Rightarrow t = -1\) \(\bar{p} = \bar{i} + (-3-1)\bar{j} - 2(-1)\bar{k} = \bar{i} - 4\bar{j} + 2\bar{k}\)…
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