Q220 / 270 · NDA 2025 (Phase 2) · Mathematics
A force $\vec{F} = 2\hat{i} - \lambda\hat{j} + 5\hat{k}$ is applied at the point $A(1, 2, 5)$. If its moment about the point $B(-1, -2, 3)$ is $16\hat{i} - 6\hat{j} + 2\lambda\hat{k}$, then what is the value of $\lambda$?
- Correct answerA$-2$
- B$0$
- C$1$
- D$2$
Solution
The position vector of the point of application $A$ with respect to point $B$ is $\vec{r} = \vec{r}_A - \vec{r}_B$.
$\vec{r} = (1 - (-1))\hat{i} + (2 - (-2))\hat{j} + (5 - 3)\hat{k} = 2\hat{i} + 4\hat{j} + 2\hat{k}$
The moment of the force about point $B$ is given by $\vec{M} = \vec{r} \times \vec{F}$.
$\vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 2 \\ 2 & -\lambda & 5 \end{vmatrix}$
$\vec{M} = \hat{i}(20 + 2\lambda) - \hat{j}(10 - 4) + \hat{k}(-2\lambda - 8)$
$\vec{M} = (20 + 2\lambda)\hat{i} - 6\hat{j} - (2\lambda + 8)\hat{k}$
It is given that $\vec{M} = 16\hat{i} - 6\hat{j} + 2\lambda\hat{k}$.
Comparing the coefficients of $\hat{i}$ and $\hat{k}$:
$20 + 2\lambda = 16 \Rightarrow 2\lambda = -4 \Rightarrow \lambda = -2$
$-2\lambda - 8 = 2\lambda \Rightarrow 4\lambda = -8 \Rightarrow \lambda = -2$
Both equations yield $\lambda = -2$.
Answer: $-2$
$\vec{r} = (1 - (-1))\hat{i} + (2 - (-2))\hat{j} + (5 - 3)\hat{k} = 2\hat{i} + 4\hat{j} + 2\hat{k}$
The moment of the force about point $B$ is given by $\vec{M} = \vec{r} \times \vec{F}$.
$\vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 2 \\ 2 & -\lambda & 5 \end{vmatrix}$
$\vec{M} = \hat{i}(20 + 2\lambda) - \hat{j}(10 - 4) + \hat{k}(-2\lambda - 8)$
$\vec{M} = (20 + 2\lambda)\hat{i} - 6\hat{j} - (2\lambda + 8)\hat{k}$
It is given that $\vec{M} = 16\hat{i} - 6\hat{j} + 2\lambda\hat{k}$.
Comparing the coefficients of $\hat{i}$ and $\hat{k}$:
$20 + 2\lambda = 16 \Rightarrow 2\lambda = -4 \Rightarrow \lambda = -2$
$-2\lambda - 8 = 2\lambda \Rightarrow 4\lambda = -8 \Rightarrow \lambda = -2$
Both equations yield $\lambda = -2$.
Answer: $-2$
Hence, the correct answer is option: 'A'.