Let the foot of perpendicular of the point P(3,−2,−9) on the plane passing through the points (−1,−2,−3),(9,3,4),(9,−2,1) be Q(α,β,γ). Then the distance of Q from the origin is :
Solutions · 1
Official solution
The equation of the plane passing through points A(−1,−2,−3), B(9,3,4), and C(9,−2,1) can be written using the determinant :
\[
\left|\[x+11010y+250z+374\]\right|=0
\]
Expanding the determinant, we get :
2x+3y−5z−7=0
Next, we find the foot of the perpendicular from point P(3,−2,−9) to the plane. Using the coordinates of P and the equation of the plane, we can find the ratio of the perpendicular distance to the sum of the squares of the coefficients of x, y, and z :
\[
\frac{2(3) + 3(-2) - 5(-9) - 7}{2^2 + 3^2 + (-5)^2} = \frac{-38}{38}
\]
We can now find the coordinates of the foot of the perpendicular, Q(α,β,γ) :
\[
\frac{\alpha - 3}{2} = \frac{\beta + 2}{3} = \frac{\gamma + 9}{-5} = -\frac{38}{38}
\]
Solving for α, β, and γ :
\[
\alpha = 3 - 2\left(-\frac{38}{38}\right) = 1
\]
\[
\beta = -2 + 3\left(-\frac{38}{38}\right) = -5
\]
\[
\gamma = -9 - 5\left(-\frac{38}{38}\right) = -4
\]
So, the coordinates of the foot of the perpendicular are Q(1,−5,−4). Now, we can find the distance of point Q from the origin :
\[
OQ = \sqrt{\alpha^2 + \beta^2 + \gamma^2} = \sqrt{1^2 + (-5)^2 + (-4)^2} = \sqrt{42}
\]