The vector equation of line passing through (-1, 3, -2) and perpendicular to the lines (x+4)/1 = y/2 = (z-3)/3 and (x+2)/(-3) = (y+5)/2 = (z-6)/5 is Option 1: r = (-3i + 4j + 15k) + (-i + 3j - 2k) Option 2: r = (-i + 3j - 2k) + (-3i + 4j + 15k) Option 3: r = 2i - 7j + 4k + (-i + 3j - 2k) Option 4: r = (-i + 3j - 2k) + (2i - 7j + 4k)

Answer: r = (-i + 3j - 2k) + (2i - 7j + 4k). Solution: Solution Coming Soon

Geometry

3D Geometry

Medium

core

The vector equation of line passing through $(-1, 3, -2)$ and perpendicular to the lines $\frac{x+4}{1} = \frac{y}{2} = \frac{z-3}{3}$ and $\frac{x+2}{-3} = \frac{y+5}{2} = \frac{z-6}{5}$ is

$\vec{r} = (-3\hat{i} + 4\hat{j} + 15\hat{k}) + \lambda(-i + 3\hat{j} - 2\hat{k})$

$\vec{r} = (-\hat{i} + 3\hat{j} - 2\hat{k}) + \lambda(-3\hat{i} + \hat{4j} + 15\hat{k})$

$\vec{r} = 2\hat{i} - 7\hat{j} + 4\hat{k} + \lambda(-i + 3\hat{j} - 2\hat{k})$

$\vec{r} = (-i + 3\hat{j} - 2\hat{k}) + \lambda(2\hat{i} - 7\hat{j} + 4\hat{k})$

✅ Correct Option: 4

Solution Coming Soon

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