Question

The distance of the point P(3, 4, 4) from the point of intersection of the line joining the points Q(3, -4, -5) and R(2, -3, 1) and the plane $2x+y+z=7$, is equal to ________.

Hint:

The most direct way to find the intersection of a line and a plane is to express the line in parametric form and substitute the expressions for x, y, and z into the plane's equation. This yields a single equation to solve for the parameter.

Answer 1

Correct Answer: 7

Step 1: Find the equation of the line passing through points Q and R.
The direction vector of the line is $\vec{v} = \vec{R} - \vec{Q} = (2-3, -3-(-4), 1-(-5)) = (-1, 1, 6)$.
The parametric equation of the line, using point Q, can be written as:
$x = 3 - t$
$y = -4 + t$
$z = -5 + 6t$
Step 2: Find the point of intersection of this line and the plane $2x+y+z=7$.
Substitute the parametric equations of the line into the equation of the plane.
$2(3-t) + (-4+t) + (-5+6t) = 7$.
$6 - 2t - 4 + t - 5 + 6t = 7$.
Combine the constant terms and the t terms:
$(6 - 4 - 5) + (-2t + t + 6t) = 7$.
$-3 + 5t = 7$.
$5t = 10 \implies t = 2$.
Step 3: Find the coordinates of the intersection point.
Substitute the value $t=2$ back into the parametric equations of the line.
Let the intersection point be I.
$x_I = 3 - 2 = 1$.
$y_I = -4 + 2 = -2$.
$z_I = -5 + 6(2) = -5 + 12 = 7$.
So, the point of intersection is I(1, -2, 7).
Step 4: Calculate the distance between point P(3, 4, 4) and point I(1, -2, 7).
Using the distance formula in 3D:
$d = \sqrt{(x_P - x_I)^2 + (y_P - y_I)^2 + (z_P - z_I)^2}$.
$d = \sqrt{(3-1)^2 + (4-(-2))^2 + (4-7)^2}$.
$d = \sqrt{2^2 + 6^2 + (-3)^2}$.
$d = \sqrt{4 + 36 + 9} = \sqrt{49}$.
$d = 7$.