1.

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

Answer»

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, \(\vec{R}=\vec{a}+λ\vec{b}\)

∴\(x\hat{i}+y\hat{j}+z\hat{k}=-\hat{i}-8\hat{j}+5\hat{k}+λ(7\hat{i}-3\hat{j}-3\hat{k})\)

=\((-1+7λ) \hat{i}+(-8-3λ) \hat{j}+(5-3λ) \hat{j}\)

⇒\(\frac{x+1}{7}=\frac{y+8}{-3}=\frac{z-5}{-3}\).


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