Question 37

Given a rectangular prism $ABCD.EFGH$ where $AB = 6$ cm, $BC = 8$ cm, and $BF = 4$ cm. If $\alpha$ is the angle between $AH$ and $BD$ , then $\cos 2\alpha =$ ....

Explanation

Consider the following rectangular prism illustration.

Rectangular Prism

ABCD.EFGH

Visualization of the rectangular prism with given dimensions and relevant lines for calculating the angle.

Lines $AH$ and $BD$ do not intersect, so we need to shift one of them so they intersect, namely shift line $AH$ to line $BG$ (because $AH$ is parallel to $BG$ ), so the angle is between lines $BG$ and $BD$ .

\alpha = \angle(AH, BD) = \angle(BG, BD)

Determine segment lengths

Then determine the length of $\triangle BDG$

BD = \sqrt{AB^2 + AD^2} = \sqrt{6^2 + 8^2} = 10

DG = \sqrt{CD^2 + CG^2} = \sqrt{6^2 + 4^2} = \sqrt{52} = 2\sqrt{13}

BG = \sqrt{CB^2 + CG^2} = \sqrt{8^2 + 4^2} = \sqrt{80} = 4\sqrt{5}

Use the cosine rule

Use the cosine rule on $\triangle BDG$

\cos \angle DBG = \frac{BD^2 + BG^2 - DG^2}{2 \cdot BD \cdot BG}

\cos \alpha = \frac{10^2 + (4\sqrt{5})^2 - (2\sqrt{13})^2}{2(10)(4\sqrt{5})}

\cos \alpha = \frac{100 + 80 - 52}{80\sqrt{5}}

\cos \alpha = \frac{128}{80\sqrt{5}}

\cos \alpha = \frac{8}{5\sqrt{5}}

Then the value of $\cos 2\alpha$ can be calculated using the double angle formula

\cos 2\alpha = 2\cos^2 \alpha - 1

\cos 2\alpha = 2\left(\frac{8}{5\sqrt{5}}\right)^2 - 1

\cos 2\alpha = 2\left(\frac{64}{125}\right) - 1

\cos 2\alpha = \frac{128}{125} - \frac{125}{125}

\cos 2\alpha = \frac{3}{125}

Therefore, the value of $\cos 2\alpha$ is $\frac{3}{125}$ .

Command Palette

Number 37

Explanation

Determine segment lengths

Use the cosine rule