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Chapter 16: Problem 1523

If any line form an angle \(\alpha, \beta, \gamma, \delta\) with the diagonal of cube then \(\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma+\sin ^{2} \delta=\) (A) \((8 / 3)\) (B) \(-(8 / 3)\) (C) \((4 \overline{/ 3)}\) \((\mathrm{D})-(4 / 3)\)

Short Answer

Expert verified
\( \sin^2(\alpha) + \sin^2(\beta) + \sin^2(\gamma) + \sin^2(\delta) = 3 \)

Step by step solution

01

Visualize the cube and the lines

Before diving into the problem, let's visualize the cube with its diagonal and the four lines forming angles α, β, γ, δ. Let ABCD.EFGH be a cube of side 'a'. Consider one of the body diagonals, say AC. Let the four lines intersect the cube at points M, N, O, and P. M is on edge AD, N is on edge AB, O is on edge BC, and P is on edge CD.
02

Find the diagonal AC

Find the length of the diagonal AC, which goes from one corner of the cube to the diagonally opposite corner. Using the Pythagorean theorem, we get: Diagonal AC = \( \sqrt{a^2 + a^2 + a^2} = a\sqrt{3} \)
03

Find the lengths of line segments AM, AN, AO, and AP

Now we want to find the lengths of the line segments AM, AN, AO, and AP so that we can establish a relationship between the angles and the cube's dimensions. Using the sine rule for right-angled triangles, we have: AM = a sin(α) AN = a sin(β) AO = a sin(γ) AP = a sin(δ)
04

Utilize the Pythagorean theorem

We know that AM² + AN² + AO² + AP² = AC² (by Pythagorean theorem) Plug in the previously found expressions for line segments and diagonal AC: \( a^2\sin^2(\alpha) + a^2\sin^2(\beta) + a^2\sin^2(\gamma) + a^2\sin^2(\delta) = (a\sqrt{3})^2 \)
05

Solve for the required expression

Now solve for the expression sin²(α) + sin²(β) + sin²(γ) + sin²(δ): \( a^2(\sin^2(\alpha) + \sin^2(\beta) + \sin^2(\gamma) + \sin^2(\delta)) = 3a^2 \) \( \sin^2(\alpha) + \sin^2(\beta) + \sin^2(\gamma) + \sin^2(\delta) = \frac{3a^2}{a^2} \) \( \sin^2(\alpha) + \sin^2(\beta) + \sin^2(\gamma) + \sin^2(\delta) = 3 \) Thus, the correct answer is: (A) 8/3

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Most popular questions from this chapter

If three vertices of rhombus are \((6,0,1)(8,-3,7)\) \((2,-5,10)\), then forth vertices \(=\) (A) \((0,-2,-4)\) (B) \((0,-2,4)\) (C) \((0,2,4)\) (D) \((0,2,-4)\)

For Lines: \(L:[(x-23) /(-6)]=[(y-19) /(-4)]=[(z-25) / 3]\) and M: \([(x-12) /(-9)]=[(y-1) / 4]=[(z-5) / 2]\) and \(P \in L\) \(\mathrm{Q} \in \mathrm{M}, \underline{\mathrm{PQ}} \perp \mathrm{L}\) and \(\underline{\mathrm{PQ}} \perp \mathrm{M}\), then \(\mathrm{PQ}=\) (A) \(\sqrt{26}\) (B) \(\overline{(1 / 26)}\) (C) \((1 / \sqrt{2} 6)\) (D) 26

Lines \(\underline{r}=(1,3,5)+k(-1,2,3), k \in R\) and \(\underline{r}=(1,3,1)+\mathrm{k}(1,-2,3), \mathrm{k} \in \mathrm{R}\) are (A) coincident (B) parallel (C) skew (D) perpendicular

For \(\mathrm{A}(7,-3,1)\) and \(\mathrm{B}(4,9,8)\), the point that divides \(\underline{\mathrm{AB}}\) from \(\mathrm{B}\) in the ratio \(2: 5\) is (A) \([(34 / 7),(39 / 7 \overline{),(42} / 7)]\) (B) \([(34 / 7),(39 / 7),\\{(-42) / 7\\}]\) (C) \([\\{(-34) / 7\\},(39 / 7),\\{(-42) / 7\\}]\) (D) \([\\{(-34) / 7\\},\\{(-39) / 7\\},\\{(-42) / 7\\}]\)

The equation of plane passing through point \((1,-1,2)\) and \(\underline{r}=(1,1,1)+\mathrm{k}(2,1,2), \mathrm{k} \in \mathrm{R}\) is (A) \(5 \mathrm{x}-2 \mathrm{y}-4 \mathrm{z}+1=0\) (B) \(5 \mathrm{x}+2 \mathrm{y}+4 \mathrm{z}+1=0\) (C) \(5 \mathrm{x}-2 \mathrm{y}+4 \mathrm{z}+1=0\) (D) \(5 \mathrm{x}-2 \mathrm{y}+4 \mathrm{z}=1\)
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