Find the cartesian equation of the line which is passing through the point (-1, -8, 5) and parallel to the vector \(7\hat{i}-3\hat{j}-3\hat{k}\).

Category: QuestionsFind the cartesian equation of the line which is passing through the point (-1, -8, 5) and parallel to the vector \(7\hat{i}-3\hat{j}-3\hat{k}\).
Editor">Editor Staff asked 11 months ago

Find the cartesian equation of the line which is passing through the point (-1, -8, 5) and parallel to the vector \(7\hat{i}-3\hat{j}-3\hat{k}\).
 
(a) \(\frac{x+1}{-1}=\frac{y+8}{-8}=\frac{z-5}{5}\)
 
(b) \(\frac{x+1}{7}=\frac{y+8}{-3}=\frac{z-5}{-3}\)
 
(c) \(\frac{x-7}{7}=\frac{y+8}{-3}=\frac{z-3}{-3}\)
 
(d) \(\frac{x+1}{7}=\frac{y+8}{-8}=\frac{z+5}{3}\)
 
I got this question during an internship interview.
 
The question is from Three Dimensional Geometry in division Three Dimensional Geometry of Mathematics – Class 12
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1 Answers
Editor">Editor Staff answered 11 months ago

Right choice is (b) \(\frac{x+1}{7}=\frac{y+8}{-3}=\frac{z-5}{-3}\)
 
Best explanation: The position vector of the given point is \(\vec{a}=-\hat{i}-8\hat{j}+5\hat{k}\)
 
The vector which is parallel to the given line is \(7\hat{i}-3\hat{j}-3\hat{k}\)
 
We know that, \(\vec{r}=\vec{a}+λ\vec{b}\)
 
∴\(x\hat{i}+y\hat{j}+z\hat{k}=-\hat{i}-8\hat{j}+5\hat{k}+λ(7\hat{i}-3\hat{j}-3\hat{k})\)
 
=\((-1+7λ) \hat{i}+(-8-3λ) \hat{j}+(5-3λ) \hat{j}\)
 
⇒\(\frac{x+1}{7}=\frac{y+8}{-3}=\frac{z-5}{-3}\).