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Paper 1 · November 2010 · Vectors

Find the angle between the planes $r \cdot (-i + 10j - 3k) = 5$ and $r \cdot (i + j + 3k) = 6$ .

A

90^\circ

B

60^\circ

C

0^\circ

D

45^\circ

Explanation

The angle between two planes equals the angle between their normals. Here

(-1)(1)+(10)(1)+(-3)(3)=-1+10-9=0

. A zero scalar product means the normals are perpendicular, so the planes meet at

90^\circ

.

Derived from ZIMSEC Further Mathematics 9187/1, November 2010, Q14

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