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}

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This intriguing question originated from Three Dimensional Geometry topic in division Three Dimensional Geometry of Mathematics – Class 12

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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.