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Question
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°.
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