Question

If \(\left[ \begin{array}{ccc}2 & 1 & -1 \\ -1 & 2 & 1 \\ 1 & -1 & 2\end{array} \right]\left[ \begin{array}{c}a \\ b \\ c\end{array} \right] = \left[ \begin{array}{c}4 \\ 0 \\ -2\end{array} \right]\) then the distance of point \(P(a,b,c)\) from the plane \(2x+y+2z = 10\) is

• A. \(2\) units.
• B. \(3\) units.
• C. \(4\) units.
• D. \(5\) units.

Hint:

First solve the three equations to get P = (1, 1, -1), then use the point-to-plane distance formula.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The matrix equation gives three linear equations in \(a, b, c\). We first find the point \(P(a,b,c)\), then find its distance from the plane \(2x+y+2z=10\).

Step 2: Write the equations:
Multiplying out the matrix gives \[ 2a + b - c = 4 \quad (1) \] \[ -a + 2b + c = 0 \quad (2) \] \[ a - b + 2c = -2 \quad (3) \]

Step 3: Solve the system:
Add (1) and (2) to remove \(c\): \(a + 3b = 4\). Add (2) and (3): \(b + 3c = -2\). From \(a = 4 - 3b\) and \(c = (-2 - b)/3\), put both into (2): \(-(4 - 3b) + 2b + (-2 - b)/3 = 0\). Multiply by 3 to get \(14b = 14\), so \(b = 1\), \(a = 1\), \(c = -1\). Check in (1): \(2 + 1 + 1 = 4\). Check in (3): \(1 - 1 - 2 = -2\). So \(P = (1, 1, -1)\).

Step 4: Distance formula:
The distance of \((x_1,y_1,z_1)\) from \(Ax + By + Cz + D = 0\) is \[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \] The plane is \(2x + y + 2z - 10 = 0\).

Step 5: Substitute:
\[ d = \frac{|2(1) + 1(1) + 2(-1) - 10|}{\sqrt{4 + 1 + 4}} = \frac{|-9|}{3} = 3 \]

Step 6: Why the others are wrong:
The numerator 9 divided by 3 gives exactly 3. The values 2, 4 and 5 would come from slips, such as forgetting the -10 or using \(c = 1\) instead of \(-1\).

Final Answer:
The point is \(P(1,1,-1)\) and its distance from the plane is 3 units, option (B). \[ \boxed{3\text{ units}} \]