Prove that the planes $x + 4 y - 3 = 1$ $x+4y-3=1$ and $- 3 x + 6 y + 7 z = 0$ $-3x+6y+7z=0$ are perpendicular to each other?

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Prove that the planes $x + 4 y - 3 = 1$ $x+4y-3=1$ and $- 3 x + 6 y + 7 z = 0$ $-3x+6y+7z=0$ are perpendicular to each other?

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Answer 1 · 2017-05-27T05:59:55

Planes are perpendicular to each other.

Explanation:

The normal to a plane $a x + b y + c z + d = 0$ $ax+by+cz+d=0$ is $< a, b, c >$ $<a,b,c>$

hence normal to $x + 4 y - 3 = 1$ $x+4y-3=1$ is $< 1, 4, - 3 >$ $<1,4,-3>$

and normal to $- 3 x + 6 y + 7 z = 0$ $-3x+6y+7z=0$ is $< - 3, 6, 7 >$ $<-3,6,7>$

and angle between planes is equal to angle between their normals.

As normals are $< 1, 4,, - 3 >$ $<1,4,,-3>$ and $< - 3, 6, 7 >$ $<-3,6,7>$ and if angle between them is $α$ $alpha$

then $cos α = \frac{| n_{1} . n_{2} |}{| n_{1} | | n_{2} |}$ $cosalpha={|n_1.n_2|)/(|n_1||n_2|}$

= $\frac{| 1 \times (- 3) + 4 \times 6 + (- 3) \times 7 |}{\sqrt{1^{2} + 4^{2} + {(- 3)}^{2}} \sqrt{{(- 3)}^{2} + 6^{2} + 7^{2}}}$ $(|1xx(-3)+4xx6+(-3)xx7|)/(sqrt(1^2+4^2+(-3)^2)sqrt((-3)^2+6^2+7^2)$

= $\frac{| - 3 + 24 - 21 |}{\sqrt{1 + 16 + 9} \sqrt{9 + 36 + 49} ∣ ∣}$ $(|-3+24-21|)/(sqrt(1+16+9)sqrt(9+36+49)|)$

= $0$ $0$

Hence. $α = 90^{\circ}$ $alpha=90^@$ and planes are perpendicular to each other.

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Prove that the planes $x + 4 y - 3 = 1$ $x+4y-3=1$ and $- 3 x + 6 y + 7 z = 0$ $-3x+6y+7z=0$ are perpendicular to each other?

1 Answer

Explanation:

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Prove that the planes $x + 4 y - 3 = 1$ $x+4y-3=1$ and $- 3 x + 6 y + 7 z = 0$ $-3x+6y+7z=0$ are perpendicular to each other?