Question

For \( a \in \mathbb{R} \), 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 and \( 3\left(\vec{p} \cdot \vec{q} \right)^2 - \lambda \left|\vec{r} \times \vec{q} \right|^2 = 0 \), then the value of \( \lambda \) is:

• A. \( \dfrac{2}{3} \)
• B. \( \dfrac{3}{2} \)
• C. \( 2 \)
• D. \( 1 \)

Hint:

To verify coplanarity conditions involving dot and cross products, simplify algebraically and equate expressions as given in the condition.

Answer 1

The Correct Option is D

Step 1: Compute \( \vec{p} \cdot \vec{q} \) \[ \vec{p} \cdot \vec{q} = [(a+1)\hat{i} + a\hat{j} + a\hat{k}] \cdot [a\hat{i} + (a+1)\hat{j} + a\hat{k}] \] \[ = (a+1)(a) + a(a+1) + a(a) = a(a+1) + a(a+1) + a^2 = 2a(a+1) + a^2 \] \[ = 2a^2 + 2a + a^2 = 3a^2 + 2a \] \[ \Rightarrow (\vec{p} \cdot \vec{q})^2 = (3a^2 + 2a)^2 \] Step 2: Compute \( |\vec{r} \times \vec{q}| \) 

Step 3: Use given equation and solve for \( \lambda \) \[ 3(\vec{p} \cdot \vec{q})^2 - \lambda |\vec{r} \times \vec{q}|^2 = 0 \] \[ \Rightarrow 3(3a^2 + 2a)^2 - \lambda (6a^2 + 4a + 1) = 0 \] \[ (3a^2 + 2a)^2 = 9a^4 + 12a^3 + 4a^2 \Rightarrow 3(9a^4 + 12a^3 + 4a^2) = 27a^4 + 36a^3 + 12a^2 \] Now compute the RHS: \[ \lambda (6a^2 + 4a + 1) \] Equating both: \[ 27a^4 + 36a^3 + 12a^2 = \lambda (6a^2 + 4a + 1) \] We can factor: \[ \lambda = \dfrac{27a^4 + 36a^3 + 12a^2}{6a^2 + 4a + 1} \] Try polynomial division: \[ (6a^2 + 4a + 1) \times 1 = 6a^2 + 4a + 1 \] \[ (6a^2 + 4a + 1) \times a = 6a^3 + 4a^2 + a \] \[ (6a^2 + 4a + 1) \times 4a = 24a^3 + 16a^2 + 4a \] \[ (6a^2 + 4a + 1) \times 6a^2 = 36a^4 + 24a^3 + 6a^2 \] So we can verify: \[ \lambda = 1 \]