Mathematics: 1. Find symmetric equations for the line that passes through the point (4, −4, 8) and is parallel to the vector −1, 4, −3 (b) Find the points in which the required line in part (a) intersects the coordinate planes. point of intersection with xy-plane point of intersection with yz-plane point of intersection with xz-plane 2. Find an equation for the plane consisting of all points that are equidistant from the points (−6, 4, 1) and (2, 6, 5). 3. Find an equation of the plane. The plane that passes

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1. Find symmetric equations for the line that passes

Ответ:

misskylie2003

14.03.2021

Mathematics

The answer is below

Step-by-step explanation:

a) The symmetric equations for the line that passes through the point (a, b, c) and is parallel to the vector (e, f, g) is:

$\frac{x-a}{e}=\frac{y-b}{f} =\frac{z-c}{g}$

Therefore using the above equation to Find symmetric equations for the line that passes through the point (4, −4, 8) and is parallel to the vector (−1, 4, −3) we get:

$\frac{x-4}{-1}=\frac{y-(-4)}{4}=\frac{z-8}{-3} \\\\Therefore:\\\\\frac{x-4}{-1}=\frac{y+4}{4}=\frac{z-8}{-3}$

b)i) The line would intersect the xy plane where z = 0. Hence:

$\frac{x-4}{-1}=\frac{y+4}{4}=\frac{0-8}{-3}\\\\\frac{x-4}{-1}=\frac{y+4}{4}=\frac{8}{3} \\\\\frac{x-4}{-1}=\frac{8}{3}\ and\ \frac{y+4}{4}=\frac{8}{3}\\\\x=\frac{4}{3}\ and\ y=\frac{20}{3} \\\\Therefore\ the\ line\ intersect\ the\ xy\ plane\ at\ (\frac{4}{3},\frac{20}{3},0)$

ii) The line would intersect the yz plane where x = 0. Hence:

$\frac{0-4}{-1}=\frac{y+4}{4}=\frac{z-8}{-3}\\\\\frac{y+4}{4}=\frac{z-8}{-3}=4\\\\\frac{y+4}{4}=4\ and\ \frac{z-8}{-3}=4\\\\y=12\ and\ z=-4 \\\\Therefore\ the\ line\ intersect\ the\ yz\ plane\ at\ (0,12,-4)$

iii) The line would intersect the xz plane where y = 0. Hence:

$\frac{x-4}{-1}=\frac{0+4}{4}=\frac{z-8}{-3}\\\\\frac{x-4}{-1}=\frac{z-8}{-3}=1\\\\\frac{x-4}{-1}=1\ and\ \frac{z-8}{-3}=1\\\\x=3\ and\ z=5 \\\\Therefore\ the\ line\ intersect\ the\ xz\ plane\ at\ (3,0,5)$

Let the points be A(−6, 4, 1) and B(2, 6, 5). Let O be the midpoint of the two points A and B. Therefore O is the average of the x coordinates, y coordinates and z coordinate.

$0=\frac{1}{2} (-6+2,4+6,1+5)=(- 2,5,3)\\\\$

The normal vector (n) in the direction of line between A and B is:

n = AB = B - A = (2, 6, 5) - (−6, 4, 1) = (8, 2, 4)

n = 8x + 2y + 4z

The equation of the plane based on the normal vector and the midpoint 0 is:

Plane = 8x + 2y + 4z = 8(-2) + 2(5) + 4(3) = 6

Therefore:

8x + 2y + 4z = 6

4x + y + 2z = 3

3) The normal vectors to the plane are:

$n_1=(1,1,-1)\ and\ n_2=(4.-1,5)$

The point of intersection O of the two planes is normal to the normal vectors, hence:

$O=n_1*n_2=(1,1,-1)*(4,-1,5)\\\\O=(4, -9, -5)$

A point that lies on both plane is gotten by substituting z = 0, hence:

x + y - (0) = 3, and 4x - y + 5(0) = 5

x + y = 3 and 4x - y = 5

Solving simultaneously gives x = 8/5, y=7/5

From this two points we get:

AB = (-2-8/5, 1 - 7/5, 1-0) = (-18/5, -2/5, 1)

The vector normal to the plane (n) = (4, -9, -5) * (-18/5, -2/5, 1) = (-11, 14, -34)

$B=A_o=-11(x-(-2))+14(y-1)-34(z-1)\\\\0=11x+22+14y-14-34z+34\\\\11x+14y-34z =42\\$

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