Question

Find the point on the plane x - 2y + 3z = 6 that is closest to the point ( 0 , 1 , 1 ) .

Answer 1

The point on the plane
`\(x - 2y + 3z = 6\)` closest to the point
`\((0, 1, 1)\) is \((2, 0, 2)\).`

Step-by-step explanation:

To find the point on the plane closest to a given point, we use the formula for the distance from a point to a plane. The formula is given by
`\(D = (\lvert Ax_0 + By_0 + Cz_0 + D \rvert)/(√(A^2 + B^2 + C^2))\), where \((x_0, y_0, z_0)\) is the point, and \(Ax + By + Cz + D = 0\)` is the equation of the plane. In this case, the point is
`\((0, 1, 1)\)`, and the plane is
`\(x - 2y + 3z = 6\).` Plugging these values into the formula, we get
`\(D = (\lvert 0 - 2(1) + 3(1) - 6 \rvert)/(√(1^2 + (-2)^2 + 3^2)) = (1)/(√(14))\).`

Next, to find the coordinates of the closest point, we use the normal vector of the plane, which is
`\((1, -2, 3)\). We scale this vector by \(D\) and add it to the given point: \((0, 1, 1) + (1)/(√(14))(1, -2, 3) = (2, 0, 2)\). Therefore, the point on the plane closest to \((0, 1, 1)\) is \((2, 0, 2)\)`. This point minimizes the distance between the plane and the given point.

Answer 2

The equation of the line BC is y = -7/5x + 48/5

To find the equation of BC, we can follow these steps:

Slope of AB = (4 - (-1)) / (4 - (-3)) = 5/7.

Since BC is perpendicular to AB, its slope will be the negative reciprocal of 5/7: -7/5.

Use the point-slope form of the equation for BC:

We know that BC passes through point B (4, 4) and has a slope of -7/5. The point-slope form of the equation is:

y - y1 = m(x - x1)

where y is the y-coordinate of any point on BC, y1 is the y-coordinate of point B (4), m is the slope (-7/5), x is the x-coordinate of any point on BC, and x1 is the x-coordinate of point B (4).

Plug in the values and solve for y:

y - 4 = (-7/5)(x - 4)

y = -7/5x + 28/5 + 4

y = -7/5x + 28/5 + 20/5

y = -7/5x + 48/5