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Question

Mathematics Question on introduction to three dimensional geometry

If A(6,3,2),B(5,1,4),C(3,−4,7),D(0,2,5)A(6, 3, 2), B(5, 1, 4), C(3, -4, 7), D(0, 2, 5) be four points, the projection of segment CDCD on the line AB AB is

A

−313- \frac {3} {13}

B

−137- \frac {13} {7}

C

−133- \frac {13} {3}

D

−713- \frac {7} {13}

Answer

−133- \frac {13} {3}

Explanation

Solution

CD→=(0−3,2+4,5−7)=−3i^+6j^−2k^\overrightarrow{CD} =(0 - 3, 2 + 4, 5 - 7) = - 3\hat{i} + 6\hat{j} - 2\hat{k} AB→=(5−6,1−3,4−2)=−i^−2j^+2k^\overrightarrow{AB} =(5 - 6, 1 - 3, 4 - 2) = - \hat{i} - 2\hat{j} + 2\hat{k} ∣AB→∣=1+4+4=3|\overrightarrow{AB}| = \sqrt{1 + 4 +4} = 3 ∴\therefore A unit vector along AB = 13(−i^−2 j^+2 k^)\frac{1}{3} (- \hat{i} - 2 \,\hat{j} + 2\, \hat{k}) ∴\therefore Projection of CD→\overrightarrow{CD} on AB→\overrightarrow{AB} = (−3i^+6j^−2k^)13(−i^−2j^+2k^) (- 3\hat{i} + 6\hat{j} - 2\hat{k}) \frac{1}{3} (- \hat{i} - 2 \hat{j} + 2 \hat{k}) = 13(3−12−4)=−133\frac{1}{3} (3 - 12 - 4) = - \frac{13}{3}