In-a-cube-ABCD-EFGH-with-a-side-length-of-4-cm-Point-Q-is-the-middle-of-line-AE-and-point-P-lies-on-the-extension-of-line-AB-so-that-BP-AB-1-2-If-is-the-angle-between-the-QG-line-and-the-PQH

Question Number 134428 by liberty last updated on 03/Mar/21

In a cube ABCD . EFGH with a

side length of 4 cm . Point Q is the

middle of line AE and point P

lies on the extension of line AB

so that BP : AB = 1 : 2 . If θ is the

angle between the QG line and

the PQH plane, then \sin θ = ?

Answered by EDWIN88 last updated on 03/Mar/21

let Q (4, 0, 2), P (4, 6, 0) and H (0, 0, 4)

\vec{n} = QH \times QP = | \begin{matrix} - 4 0 2 \\ 0 6 - 2 \end{matrix} | = - 12 \hat{i} - 8 \hat{j} - 24 \hat{k}

Let QX = (x - 4, y, z - 2) and QX ⊥ \vec{n}

we get the equation of plane PQH is

\Rightarrow - 12 (x - 4) - 8 y - 24 (z - 2) = 0

\Rightarrow - 12 x - 8 y - 24 z + 96 = 0 or

\Rightarrow 3 x + 2 y + 6 z - 24 = 0

Let d be a distance a point G (0, 4, 4) to PQH plane

then we get d = \frac{∣ 8 + 24 - 24 ∣}{\sqrt{9 + 4 + 36}} = \frac{8}{7}

so \sin θ = \frac{d}{∣ QG ∣} = \frac{\frac{8}{7}}{\sqrt{16 + 16 + 4}} = \frac{8}{7 \times 6}

\sin θ = \frac{4}{21} .

Answered by mr W last updated on 03/Mar/21

e q n . o f P Q H :

\frac{x}{4} - \frac{y}{6} + \frac{z}{2} = 1

\boldsymbol n = (\frac{1}{4}, - \frac{1}{6}, \frac{1}{2})

\boldsymbol G Q = (- 4, - 4, 2)

\sin θ = \cos (90 - θ) = \frac{\boldsymbol n \cdot \boldsymbol G Q}{∣ \boldsymbol n ∣∣ \boldsymbol G Q ∣}

= \frac{- 4 \times \frac{1}{4} - 4 \times (- \frac{1}{6}) + 2 \times \frac{1}{2}}{\sqrt{(\frac{1}{4^{2}} + \frac{1}{8^{2}} + \frac{1}{2^{2}}) (4^{2} + 4^{2} + 2^{2}})} = \frac{4}{21}

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