Find the angle between the line $\frac{x - 2}{3} = \frac{y + 3}{3} = \frac{z - 1}{1}$ $(x-2)/3=(y+3)/3=(z-1)/1$ and the plane $2 x - 3 y + 4 z = 0$ $2x-3y+4z=0$ ?

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Find the angle between the line $\frac{x - 2}{3} = \frac{y + 3}{3} = \frac{z - 1}{1}$ $(x-2)/3=(y+3)/3=(z-1)/1$ and the plane $2 x - 3 y + 4 z = 0$ $2x-3y+4z=0$ ?

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Answer 1 · 2018-06-22T15:44:43

look carefully, $θ$ $theta$ is the anlge b/w plane and line.

Explanation:

given the direction cosine of line as $\to b = 3 ˆ i + 3 ˆ j + ˆ k$ $vec b = 3hati+3hatj+hatk$ and that of plane as $\to n = 2 ˆ i - 3 ˆ j + 4 ˆ k$ $vec n = 2hati-3hatj+4hatk$
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clearly , using dot product of vectors,
$cos (90 - θ) = ˆ b . ˆ n$ $cos(90-theta)=hatb.hatn$
Nowit is very easy to find the numerical value of $θ$ $theta$

Answer 2 · 2018-06-22T16:01:13

$θ \approx 0.0565 rad ({3.238}^{\circ})$ $theta ~~ 0.0565" rad "(3.238^@)$

Explanation:

We are given the symmetric form of the equation of the line:

$\frac{x - 2}{3} = \frac{y + 3}{3} = \frac{z - 1}{1}$ $(x-2)/3=(y+3)/3=(z-1)/1$

Convert the symmetric form of the line to the parametric form by setting each expression equal to t:

$\frac{x - 2}{3} = t$ $(x-2)/3= t$

$\frac{y + 3}{3} = t$ $(y+3)/3=t$

$\frac{z - 1}{1} = t$ $(z-1)/1= t$

Multiply both sides by the denominators:

$x - 2 = 3 t$ $x-2= 3t$

$y + 3 = 3 t$ $y+3=3t$

$z - 1 = t$ $z-1= t$

Move the constants to the right:

$x = 3 t + 2$ $x= 3t+2$

$y = 3 t - 3$ $y=3t-3$

$z = t + 1$ $z= t+1$

Convert to the vector form:

$(x, y, z) = (2, - 3, 1) + t (3 ˆ i + 3 ˆ j + ˆ k)$ $(x,y,z) = (2,-3,1)+ t(3hati+3hatj+hatk)$

Please understand the $\to u = 3 ˆ i + 3 ˆ j + ˆ k$ $vecu = 3hati+3hatj+hatk$ is the direction of the line.

Find any two points in the plane, $2 x - 3 y + 4 z = 0$ $2x-3y+4z=0$ :

Pick $(0, 0, ?)$ $(0,0,?)$ :

$2 (0) - 3 (0) + 4 z = 0$ $2(0)-3(0)+4z=0$

$z = 0$ $z = 0$

The first point is $(0, 0, 0)$ $(0,0,0)$

Pick $(4, 4, ?)$ $(4,4,?)$

$2 (4) - 3 (4) + 4 z = 0$ $2(4)-3(4)+4z=0$

$z = 1$ $z = 1$

The second point is $(4, 4, 1)$ $(4,4,1)$

Make a vector from the first point to the second:

$\to v = (4 - 0) ˆ i + (4 - 0) ˆ j + (1 - 0) ˆ k$ $vecv = (4-0)hati+(4-0)hatj+(1-0)hatk$

$\to v = 4 ˆ i + 4 ˆ j + 1 ˆ k$ $vecv = 4hati+4hatj+1hatk$

Please understand that $\to v$ $vecv$ is in the plane and the angle between $\to u$ $vecu$ and $\to v$ $vecv$ is, also, the angle between the line and the plane.

Compute the dot-product of $\to u$ $vecu$ and $\to v$ $vecv$ :

$\to u \cdot \to v = 3 (4) + 3 (4) + 1 (1) = 25$ $vecu*vecv = 3(4)+3(4)+1(1)=25$

Compute the magnitudes of $\to u$ $vecu$ and $\to v$ $vecv$ :

$∣ ∣ \to u ∣ ∣ = \sqrt{3^{2} + 3^{2} + 1^{2}} = \sqrt{19}$ $|vecu|= sqrt(3^2+3^2+1^2) = sqrt19$

$∣ ∣ \to v ∣ ∣ = \sqrt{4^{2} + 4^{2} + 1^{2}} = \sqrt{33}$ $|vecv|= sqrt(4^2+4^2+1^2) = sqrt33$

Using the formula $\to u \cdot \to v = ∣ ∣ \to u ∣ ∣ ∣ ∣ \to v ∣ ∣ cos (θ)$ $vecu*vecv=|vecu||vecv|cos(theta)$ , we can find the angle between the two vectors:

$25 = \sqrt{19} \sqrt{33} cos (θ)$ $25 = sqrt19sqrt33cos(theta)$

$θ = {cos}^{- 1} (\frac{25}{\sqrt{19} \sqrt{33}})$ $theta = cos^-1(25/(sqrt19sqrt33))$

$θ \approx 0.0565 rad ({3.238}^{\circ})$ $theta ~~ 0.0565" rad "(3.238^@)$

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Find the angle between the line $\frac{x - 2}{3} = \frac{y + 3}{3} = \frac{z - 1}{1}$ $(x-2)/3=(y+3)/3=(z-1)/1$ and the plane $2 x - 3 y + 4 z = 0$ $2x-3y+4z=0$ ?

2 Answers

Explanation:

Explanation:

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Find the angle between the line $\frac{x - 2}{3} = \frac{y + 3}{3} = \frac{z - 1}{1}$ $(x-2)/3=(y+3)/3=(z-1)/1$ and the plane $2 x - 3 y + 4 z = 0$ $2x-3y+4z=0$ ?