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InterviewSolution
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Find the cartesian equation of the line which is passing through the point (5,-6,1) and parallel to the vector \(5\hat{i}+2\hat{j}-3\hat{k}\).(a) \(\frac{x-5}{5}=\frac{y-4}{6}=\frac{z+1}{-3}\)(b) \(\frac{x-5}{5}=\frac{z+6}{2}=\frac{y-1}{3}\)(c) \(\frac{x+5}{4}=\frac{y-8}{2}=\frac{z-1}{-3}\)(d) \(\frac{x-5}{5}=\frac{y+6}{2}=\frac{z-1}{-3}\)This question was addressed to me during an online interview.This intriguing question originated from Three Dimensional Geometry topic in division Three Dimensional Geometry of Mathematics – Class 12 |
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Answer» CORRECT choice is (d) \(\frac{x-5}{5}=\frac{y+6}{2}=\frac{z-1}{-3}\) Explanation: The equation of a line passing through a point and parallel to a VECTOR is GIVEN by \(\vec{r}=\vec{a}+λ\vec{B}\) \(\vec{a}\) is the position vector of the given point ∴\(\vec{a}=5\hat{i}-6\hat{j}+\hat{k}\) \(\vec{b}=5\hat{i}+2\hat{j}-3\hat{k}\). \(\vec{r}=5\hat{i}-6\hat{j}+\hat{k}+λ(5\hat{i}+2\hat{j}-3\hat{k})\) \(x\hat{i}+y\hat{j}+z\hat{k}=(5+5λ) \hat{i}+(2λ-6) \hat{j}+(1-3λ) \hat{k}\) ∴\(\frac{x-5}{5}=\frac{y+6}{2}=\frac{z-1}{-3}\) is the cartesian equation of the given line. |
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