KCET · Maths · Matrices
The angle between two diagonals of a cube is \( 30^{\circ} \)
- A \( 30^{\circ} \)
- B \( 45^{\circ} \)
- C \( \cos ^{-1}\left(\frac{1}{3}\right) \)
- D \( \cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \)
Answer & Solution
Correct Answer
(C) \( \cos ^{-1}\left(\frac{1}{3}\right) \)
Step-by-step Solution
Detailed explanation
Consider a diagonal with each side \(1.\)
Now, \( B C \) and \( O A \) are diagonals.
So, Angle between diagonals = Angle between
\( \overline{\mathrm{OA}} \) and \( \overline{\mathrm{BC}} \)
\( \overline{\mathrm{OA}}=(1,1,1)-(0,0,0)=(1,1,1) \) \( \overline{\mathrm{BC}}=(0,1,1)-(1,0,0)=(-1,1,1) \)
Now,
\(
\begin{array}{l}
\cos \theta=\frac{1(-1)+1(1)+(1)(1)}{\sqrt{1^{2}+1^{2}+1^{2}} \sqrt{(-1)^{2}+1^{2}+1^{2}}}=\frac{1}{3} \\
\Rightarrow \theta=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)
\end{array}
\)
Now, \( B C \) and \( O A \) are diagonals.
So, Angle between diagonals = Angle between
\( \overline{\mathrm{OA}} \) and \( \overline{\mathrm{BC}} \)
\( \overline{\mathrm{OA}}=(1,1,1)-(0,0,0)=(1,1,1) \) \( \overline{\mathrm{BC}}=(0,1,1)-(1,0,0)=(-1,1,1) \)
Now,
\(
\begin{array}{l}
\cos \theta=\frac{1(-1)+1(1)+(1)(1)}{\sqrt{1^{2}+1^{2}+1^{2}} \sqrt{(-1)^{2}+1^{2}+1^{2}}}=\frac{1}{3} \\
\Rightarrow \theta=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)
\end{array}
\)
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