Question

The values of a so that the vectors \(a\hat{i}+(a-1)\hat{j}+3\hat{k}\) and \((a+2)\hat{i}+a\hat{j}-2\hat{k}\) are perpendicular, are

• A. \(\frac{3}{2},-2\)
• B. \(2,\frac{3}{2}\)
• C. \(-2,\frac{-3}{2}\)
• D. \(2,\frac{-3}{2}\)
• E. \(-4,\frac{3}{2}\)

Answer 1

The Correct Option is A

Step 1: Use the condition for perpendicular vectors  
Two vectors \( \overrightarrow{A} \) and \( \overrightarrow{B} \) are perpendicular if their dot product is zero: \[ \overrightarrow{A} \cdot \overrightarrow{B} = 0 \] Given vectors: \[ \overrightarrow{A} = a\hat{i} + (a-1)\hat{j} + 3\hat{k} \] \[ \overrightarrow{B} = (a+2)\hat{i} + a\hat{j} -2\hat{k} \]

Step 2: Compute the dot product 
Using the formula: \[ \overrightarrow{A} \cdot \overrightarrow{B} = a(a+2) + (a-1)(a) + (3)(-2) \] Expanding: \[ a^2 + 2a + a^2 - a - 6 = 0 \] \[ 2a^2 + a - 6 = 0 \]

Step 3: Solve the quadratic equation 
The quadratic equation is: \[ 2a^2 + a - 6 = 0 \] Using the quadratic formula: \[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 2 \), \( b = 1 \), and \( c = -6 \): \[ a = \frac{-1 \pm \sqrt{(1)^2 - 4(2)(-6)}}{2(2)} \] \[ = \frac{-1 \pm \sqrt{1 + 48}}{4} \] \[ = \frac{-1 \pm \sqrt{49}}{4} \] \[ = \frac{-1 \pm 7}{4} \] \[ a = \frac{-1 + 7}{4} = \frac{6}{4} = \frac{3}{2}, \quad a = \frac{-1 - 7}{4} = \frac{-8}{4} = -2 \]

Final Answer: \( a \) is \(\frac{3}{2}, -2\).