The length of the projection of the line segment joining the points (5, –1, 4) and (4,

Engineering

Mathematics

Distance from a Plane

Question

The length of the projection of the line segment joining the points (5, –1, 4) and (4, –1, 3) on the plane, x + y + z = 7 is

$\frac{2}{\sqrt{3}}$

$\frac{2}{3}$

$\frac{1}{3}$

$\sqrt{\frac{2}{3}}$

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A(5, –1, 4), B = (4, –1, 3)

AB = $- \hat{i} - \hat{k}$

normal to plane = $\vec{n} = \hat{i} + \hat{j} + \hat{k}$

angle between line and plane is

sin θ = $\frac{\vec{AB} \cdot \vec{n}}{| \vec{AB} | | \vec{n} |} = \frac{- 1 - 1}{\sqrt{2} \sqrt{3}} = \frac{\sqrt{2}}{\sqrt{3}}$

$∴$ cos θ = $\sqrt{1 - \frac{2}{3}} = \frac{1}{\sqrt{3}}$

$∴$ projection = PQ = AB cos θ = $\frac{\sqrt{2}}{\sqrt{3}}$