# 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) $$\frac{x+1}{7}=\frac{y+8}{-8}=\frac{z+5}{3}$$

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The question is from Three Dimensional Geometry in division Three Dimensional Geometry of Mathematics – Class 12
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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, $$\vec{r}=\vec{a}+λ\vec{b}$$

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

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

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