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Find the cartesian equation of a line passing through two points (1,-9,8) and (4,-1,6).(a) \(\frac{x+1}{3}=\frac{y-9}{8}=\frac{-z-8}{2}\)(b) \(\frac{x-1}{3}=\frac{y+9}{8}=\frac{z-8}{-2}\)(c) \(\frac{x-1}{7}=\frac{y+9}{-2}=\frac{z-8}{5}\)(d) \(\frac{2x-1}{3}=\frac{6y+9}{8}=\frac{4z-8}{-2}\)I had been asked this question in class test.This intriguing question originated from Three Dimensional Geometry topic in division Three Dimensional Geometry of Mathematics – Class 12

Answer»

The CORRECT answer is (b) \(\frac{x-1}{3}=\frac{y+9}{8}=\frac{z-8}{-2}\)

Easiest explanation: The position vector for the point A(1,-9,8) and B(4,-1,6)

\(\vec{a}=\hat{i}-9\hat{j}+8\hat{k}\)

\(\vec{b}=4\hat{i}-\hat{j}+6\hat{k}\)

∴\(\vec{r}=\vec{a}+λ(\vec{b}-\vec{a})\)

The above vector equation can be expressed in CARTESIAN FORM as:

\(\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}\)

∴ The cartesian equation for the given line is \(\frac{x-1}{3}=\frac{y+9}{8}=\frac{z-8}{-2}\)


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