First slide

Direction ratios and direction cosines

Question

The angle between any two diagonals of a cube is ${Cos}^{- 1} \frac{1}{t}$ . Then value of $t^{t}$ is

Easy

A

3
B

9
C

27
D

8

Solution

$\begin{array}{l} Two diagonals OP and CS \\ \vec{OP} direction ratios are 〈1,1,1〉 \\ \overset{⇀}{CS} direction ratios are 〈1,1,-1〉 \\ if θ is the angle between two diagonals O P a n d C S t h e n \\ \cos θ = \frac{1.1 + 1.1 + 1 (1)}{\sqrt{1^{2} + 1^{2} + 1^{2}} . \sqrt{1^{2} + 1^{2} + {(- 1)}^{2}}} \\ = \frac{1 + 1 + 1}{\sqrt{3} \sqrt{3}} \\ = \frac{1}{3} \\ ⇒ θ = \cos^{- 1} (\frac{1}{3}) \\ it gives t = 3 \Rightarrow t^{t} = 3^{3} = 27 \end{array}$

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