Vector Algebra – Cross Product – JEE Main 02 April 2025 Shift 2

Solution:

$$\vec{a} = 2\hat{i} – 3\hat{j} + \hat{k}$$
$$\vec{b} = 3\hat{i} + 2\hat{j} + 5\hat{k}$$

$$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ 3 & 2 & 5 \end{vmatrix} $$
$$ \vec{a} \times \vec{b} = \hat{i}(-15 – 2) – \hat{j}(10 – 3) + \hat{k}(4 – (-9)) $$
$$ \vec{a} \times \vec{b} = -17\hat{i} – 7\hat{j} + 13\hat{k} $$

$$ (\vec{a} – \vec{c}) \times \vec{b} = -18\hat{i} – 3\hat{j} + 12\hat{k} $$
$$ (\vec{a} \times \vec{b}) – (\vec{c} \times \vec{b}) = -18\hat{i} – 3\hat{j} + 12\hat{k} $$
$$ (\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) = -18\hat{i} – 3\hat{j} + 12\hat{k} $$

Substitute $\vec{b} \times \vec{c} = \vec{d}$:
$$ (\vec{a} \times \vec{b}) + \vec{d} = -18\hat{i} – 3\hat{j} + 12\hat{k} $$
$$ \vec{d} = (-18\hat{i} – 3\hat{j} + 12\hat{k}) – (\vec{a} \times \vec{b}) $$
$$ \vec{d} = (-18\hat{i} – 3\hat{j} + 12\hat{k}) – (-17\hat{i} – 7\hat{j} + 13\hat{k}) $$
$$ \vec{d} = -\hat{i} + 4\hat{j} – \hat{k} $$

$$ \vec{a} \cdot \vec{d} = (2\hat{i} – 3\hat{j} + \hat{k}) \cdot (-\hat{i} + 4\hat{j} – \hat{k}) $$
$$ \vec{a} \cdot \vec{d} = (2)(-1) + (-3)(4) + (1)(-1) $$
$$ \vec{a} \cdot \vec{d} = -2 – 12 – 1 = -15 $$

$$ |\vec{a} \cdot \vec{d}| = |-15| = 15 $$

Ans. (4)