Question

Let P₁ be the plane 3x-y-7z = 11 and P₂ be the plane passing through the points (2,-1,0), (2,0,-1), and (5,1,1). If the foot of the perpendicular drawn from the point (7,4,-1) on the line of intersection of the planes P₁ and P₂ is (α, β, γ), then a + ẞ+ y is equal to

Hint:

To find the foot of the perpendicular from a point to a line, parameterize the line and use the condition that the vector connecting the point to the line is perpendicular to the line's direction vector.

Answer 1

Correct Answer: 11

To solve this problem, we need to find the foot of the perpendicular from the given point on the line of intersection of two planes. The first plane is defined by the equation \(3x-y-7z=11\). The second plane passes through the points \((2,-1,0)\), \((2,0,-1)\), and \((5,1,1)\). First, let's find the equation of the second plane:

• Calculate vectors from the given points: Vector 1 is from \((2,-1,0)\) to \((2,0,-1)\): \((0,1,-1)\). Vector 2 is from \((2,-1,0)\) to \((5,1,1)\): \((3,2,1)\).

Find the cross product of these vectors to get a normal vector to plane \(P_2\): \[\vec{N_2} = (0, 1, -1) \times (3, 2, 1) = (3, -3, -3)\].

Substituting point \((2,-1,0)\) into the general plane equation \(3x-3y-3z=D\), calculate \(D\):

\[3(2) - 3(-1) - 3(0) = D \Rightarrow D = 9\]. Thus, the equation of plane \(P_2\) is \(x-y-z=3\).

The line of intersection of \(P_1\) and \(P_2\) lies along their normal vector cross product:

\(\vec{N_1} = (3, -1, -7)\) and \(\vec{N_2} = (3, -3, -3)\).

\[\vec{d} = \vec{N_1} \times \vec{N_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & -7 \\ 3 & -3 & -3 \end{vmatrix} = (-30, -12, -6)\].

The direction vector of the intersection line is simplified to \(\vec{d} = (-5, -2, -1)\).

Next, find a point on this intersection line. Solve \(3x-y-7z=11\) and \(x-y-z=3\) simultaneously:

• Set \(z = 0\), and solve: From \(x-y=3\) and \(3x-y=11\), subtract to get \(2x=8 \Rightarrow x=4\), then \(y=1\).

Thus, the point \((4,1,0)\) lies on both planes and along the line of intersection.

The line equation is \((x_0,y_0,z_0) + t\vec{d} \rightarrow (4,1,0) + t(-5, -2, -1)\).

Parametric form: \((x,y,z)=(4-5t,1-2t,0-t)\).

To get the foot of the perpendicular from point \((7,4,-1)\), compute t such that the direction vector between this point and the intersection point is perpendicular to the line:

\[ \vec{n} = (7-(4-5t), 4-(1-2t), -1-(0-t))= (3+5t, 3+2t, -1+t) \]

Since \(\vec{n} \cdot \vec{d} = 0\), solve: \[(3+5t)(-5) + (3+2t)(-2) + (-1+t)(-1) = 0\]

\(-15-25t-6-4t+1-t=0\) simplify: \(-20 -30t = 0\) so \(t=-\frac{2}{3}\).

Find \((α,β,γ)\):

\[x=4-5(-\frac{2}{3}), y=1-2(-\frac{2}{3}), z=-(-\frac{2}{3})\]

which simplifies to \((α,β,γ)=(\frac{22}{3},\frac{5}{3},\frac{2}{3})\).

Finally, \(\alpha + \beta + \gamma=\frac{22}{3}+\frac{5}{3}+\frac{2}{3}=11\).

Confirm the calculated sum, \(\alpha+\beta+\gamma=11\), falls within the expected range [11,11].