Question

If the vectors \( \vec{p} = (a+1)\hat{i} + a\hat{j} + a\hat{k},\ \vec{q} = a\hat{i} + (a+1)\hat{j} + a\hat{k} \) and \( \vec{r} = a\hat{i} + a\hat{j} + (a+1)\hat{k} \) are coplanar, then the value of \(a\) is _ _ _ _.

Hint:

For symmetric vectors of this form, adding all rows simplifies the determinant quickly.

Answer 1

Correct Answer: 1

Concept: Three vectors are coplanar if their scalar triple product is zero: \[ [\vec{p}\ \vec{q}\ \vec{r}] = \vec{p} \cdot (\vec{q} \times \vec{r}) = 0 \]

Step 1: Write the determinant. \[ \begin{vmatrix} a+1 & a & a \\ a & a+1 & a \\ a & a & a+1 \end{vmatrix} = 0 \]

Step 2: Add all rows. \[ R_1 \to R_1 + R_2 + R_3 \] \[ \begin{vmatrix} 3a+1 & 3a+1 & 3a+1 \\ a & a+1 & a \\ a & a & a+1 \end{vmatrix} = 0 \]

Step 3: Factor \((3a+1)\) from \(R_1\). \[ (3a+1) \begin{vmatrix} 1 & 1 & 1 \\ a & a+1 & a \\ a & a & a+1 \end{vmatrix} = 0 \]

Step 4: Perform column operations. \[ C_2 \to C_2 - C_1,\quad C_3 \to C_3 - C_1 \] \[ (3a+1) \begin{vmatrix} 1 & 0 & 0 \\ a & 1 & 0 \\ a & 0 & 1 \end{vmatrix} = 0 \]

Step 5: Evaluate determinant. The determinant of a triangular matrix is the product of diagonal entries: \[ (3a+1) \times (1 \times 1 \times 1) = 0 \] \[ 3a + 1 = 0 \quad \Rightarrow \quad a = -\frac{1}{3} \]