# Find the angle P of the triangle whose vertices are P(0, - 1, - 2), Q(3, 1, 4) and R(5, 7, 1). - Mathematics and Statistics

Sum

Find the angle P of the triangle whose vertices are P(0, - 1, - 2), Q(3, 1, 4) and R(5, 7, 1).

#### Solution

The position vectors bar"p", bar"q", and bar"r" of the points P(0, - 1, - 2), Q(3, 1, 4) and R(5, 7, 1) are
bar"p" = - hat"j" - 2hat"k" ,
bar"q" = 3hat"i" + hat"j" + 4hat"k",
bar"r" = 5hat"i" + 7hat"j" + hat"k"

∴ bar"PQ" = bar"q" - bar"p"

= (3hat"i" + hat"j" + 4hat"k") - (- hat"j" - 2hat"k")

= 3hat"i" + 2hat"j" + 6hat"k"

and bar"PR" = bar"r" - bar"p"

= (5hat"i" + 7hat"j" + hat"k") - (- hat"j" - 2hat"k")

= 5hat"i" + 8hat"j" +3hat"k"

= bar"PQ" . bar"PR" = (3hat"i" + 2hat"j" + 6hat"k").(5hat"i" + 8hat"j" +3hat"k")

= (3)(5) + (2)(8) + (6)(3)

= 15 + 16 + 18 = 49

|bar"PQ"| = sqrt(3^2 + 2^2 + 6^2) = sqrt(9 + 4 + 36) =sqrt49 = 7

|bar"PR"| = sqrt(5^2 + 8^2 + 3^2) = sqrt(25 + 64 + 9) = sqrt98 = 7sqrt2

Using the formula for angle between two vectors,

cos P = (bar"PQ".bar"PR")/(|bar"PQ"||bar"PR"|)

= 49/(7 xx 7sqrt2) = 1/sqrt2 = "cos" 45^circ

∴ P = 45°.

Concept: Vector Product of Vectors (Cross)
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