First slide

Planes in 3D

Question

$The ratio in which the plane \vec{r} \cdot (\vec{i} - 2 \vec{j} + 3 \vec{k}) = 17 divides the line joining the points - 2 \vec{i} + 4 \vec{j} + 7 \vec{k} and$ $3 \vec{i} - 5 \vec{j} + 8 \vec{k} is$

Moderate

A

1:5
B

1:10
C

3:5
D

3:10

Solution

$Let the plane \vec{r} \cdot (\vec{i} - 2 \vec{j} + \vec{3 k}) = 17 divide the line joining$
$the points - \vec{2 i} + \vec{4} \dot{j} + \vec{7 k} and \vec{3 i} - \vec{5 j} + \vec{8 k} in the ratio t : 1 at point P$
Therefore, point P is
$\frac{3 t - 2}{t + 1} \vec{i} + \frac{- 5 t + 4}{t + 1} \vec{j} + \frac{8 t + 7}{t + 1} \vec{k}$
This lies on the given plane
$∴ \frac{3 t - 2}{t + 1} \cdot (1) + \frac{- 5 t + 4}{t + 1} (- 2) + \frac{8 t + 7}{t + 1} (3) = 17$
Solving, we get
$t = \frac{3}{10}$

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