Let P be the plane passing through the points (5, 3, 0), (13, 3, –2) and (1, 6, 2). For α ∈ N, if the distances of the points A (3, 4, α) and B (2, α, a) from the plane P are 2 and 3 respectively, then the positive value of a is
- 5
- 6
- 3
- 4
The Correct Option is D
Solution and Explanation
Step 1: Compute the normal vector of the plane
The determinant is computed to find the normal vector of the plane. Expanding along the first row:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 8 & 0 & -2 \\ 4 & -3 & -2 \end{vmatrix} = \hat{i}(-6) + 8\hat{j} - 24\hat{k} \]
Simplifying the determinant gives the normal vector:
\[ \text{Normal vector of the plane: } 3\hat{i} - 4\hat{j} + 12\hat{k} \]
Step 2: Write the equation of the plane
The equation of the plane is determined using the normal vector and a point on the plane:
\[ 3x - 4y + 12z = 3 \]
Step 3: Find \(\alpha\) (distance from \(A(3,4,\alpha)\))
The distance formula is applied to the point \(A(3,4,\alpha)\) and equated to 2:
\[ \frac{|3(3) - 4(4) + 12\alpha - 3|}{\sqrt{3^2 + (-4)^2 + 12^2}} = 2 \]
Simplify the numerator:
\[ |9 - 16 + 12\alpha - 3| = |12\alpha - 10| \]
Simplify the denominator:
\[ \sqrt{3^2 + (-4)^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 \]
The equation becomes:
\[ \frac{|12\alpha - 10|}{13} = 2 \]
Multiply through by 13:
\[ |12\alpha - 10| = 26 \]
Solving gives two possible values for \(\alpha\):
\[ 12\alpha - 10 = 26 \quad \Rightarrow \quad \alpha = 3 \] \[ 12\alpha - 10 = -26 \quad \Rightarrow \quad \alpha = -8 \]
\(\alpha = -8\) is rejected based on the given conditions or physical constraints. Thus:
\[ \alpha = 3 \]
Step 4: Find \(a\) (distance from \(B(2,3,a)\))
Similarly, the distance formula is applied to the point \(B(2,3,a)\) and equated to 3:
\[ \frac{|3(2) - 4(3) + 12a - 3|}{13} = 3 \]
Simplify the numerator:
\[ |6 - 12 + 12a - 3| = |12a - 9| \]
The equation becomes:
\[ \frac{|12a - 9|}{13} = 3 \]
Multiply through by 13:
\[ |12a - 9| = 39 \]
Solving gives two possible values for \(a\):
\[ 12a - 9 = 39 \quad \Rightarrow \quad a = 4 \] \[ 12a - 9 = -39 \quad \Rightarrow \quad a = -2.5 \]
\(a = -2.5\) is rejected based on the given conditions or physical constraints. Thus:
\[ a = 4 \]
Final Answer
The values are:
- \(\alpha = 3\)
- \(a = 4\)
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