Question

Let $\vec{u}=\hat{i}-\hat{j}-2 \hat{k}, \vec{v}=2 \hat{i}+\hat{j}-\hat{k}, \vec{v} \cdot \vec{w}=2$ and $\vec{v} \times \vec{w}=\vec{u}+\lambda \vec{v}$.  Then $\vec{u} \cdot \vec{w}$ is equal to

• A. 1
• B. 2
• C. $\frac{3}{2}$
• D. $-\frac{2}{3}$

Answer 1

The Correct Option is A

$\overrightarrow{u}=(1,-1,-2),\overrightarrow{v}=(2,1,-1),\overrightarrow{v}\cdot \overrightarrow{w}=2$
$\overrightarrow{v}\times \overrightarrow{w}=\overrightarrow{u}+\lambda \overrightarrow{v}\dots \dots \dots \dots \dots \dots \dots (1)$
Taking dot with $\overrightarrow{w}\text{in}(1)$
$\overrightarrow{w}\cdot (\overrightarrow{v}\times \overrightarrow{w})=\overrightarrow{u}\cdot \overrightarrow{w}+\lambda \overrightarrow{v}\cdot \overrightarrow{w}$
$\Rightarrow 0=\overrightarrow{u}\cdot \overrightarrow{w}+2\lambda$
Taking dot with $\vec{v}$ in (1)
$\overrightarrow{v}\cdot (\overrightarrow{v}\times \overrightarrow{w})=\overrightarrow{u}\cdot \overrightarrow{v}+\lambda \overrightarrow{v}\cdot \overrightarrow{v}$
$\Rightarrow 0=(2-1+2)+\lambda \cdot (6)$
$\lambda =-\frac{1}{2}$
$\Rightarrow \overrightarrow{u}\cdot \overrightarrow{w}=-2\lambda =1$

Answer 2

Given: \[ \mathbf{u} = (1, -1, -2), \quad \mathbf{v} = (2, 1, -1), \quad \mathbf{w} = 2. \] The vector equation is: \[ \mathbf{v} \times \mathbf{w} = \mathbf{u} + \lambda \mathbf{v} \quad \text{(1)}. \] Step 1: Dot product with \(\mathbf{w}\) Taking the dot product of both sides of equation (1) with \(\mathbf{w}\): \[ \mathbf{w} \cdot (\mathbf{v} \times \mathbf{w}) = \mathbf{w} \cdot \mathbf{u} + \lambda (\mathbf{w} \cdot \mathbf{v}). \] Using the property \(\mathbf{w} \cdot (\mathbf{v} \times \mathbf{w}) = 0\) (a vector dotted with its cross product is always zero): \[ 0 = \mathbf{w} \cdot \mathbf{u} + \lambda (\mathbf{w} \cdot \mathbf{v}). \] Substitute \(\mathbf{w} \cdot \mathbf{v} = 2\) (given): \[ 0 = \mathbf{w} \cdot \mathbf{u} + 2\lambda. \] Thus: \[ \mathbf{w} \cdot \mathbf{u} = -2\lambda. \] Step 2: Dot product with \(\mathbf{v}\) Taking the dot product of both sides of equation (1) with \(\mathbf{v}\): \[ \mathbf{v} \cdot (\mathbf{v} \times \mathbf{w}) = \mathbf{v} \cdot \mathbf{u} + \lambda (\mathbf{v} \cdot \mathbf{v}). \] Using the property \(\mathbf{v} \cdot (\mathbf{v} \times \mathbf{w}) = 0\): \[ 0 = \mathbf{v} \cdot \mathbf{u} + \lambda (\mathbf{v} \cdot \mathbf{v}). \] Substitute \(\mathbf{v} \cdot \mathbf{v} = 6\) and calculate \(\mathbf{v} \cdot \mathbf{u} = 2 \cdot 1 + 1 \cdot (-1) + (-1) \cdot (-2) = 2 - 1 + 2 = 3\): \[ 0 = 3 + 6\lambda. \] Solve for \(\lambda\): \[ \lambda = -\frac{1}{2}. \] Step 3: Calculate \(\mathbf{u} \cdot \mathbf{w}\) Substitute \(\lambda = -\frac{1}{2}\) into \(\mathbf{w} \cdot \mathbf{u} = -2\lambda\): \[ \mathbf{w} \cdot \mathbf{u} = -2 \cdot \left(-\frac{1}{2}\right) = 1. \] Thus: \[ \mathbf{u} \cdot \mathbf{w} = 1. \]