If P = 3i + √(3)j + 2k and Q = 4i + √(3)j + 2.5k then, the unit vector in the direction of P × Q is 1/x(√(3)i + j - 2√(3)k). The value of x is _________. Numerical answer type. Work it out, then open the answer.

Answer: 4. P × Q = i(2.5√(3) - 2√(3)) - j(7.5 - 8) + k(3√(3) - 4√(3)) = 1/2(√(3)i + j - 2√(3)k) Magnitude of √(3)i + j - 2√(3)k is √(3 + 1 + 12) = 4. Unit vector = 1/4(√(3)i + j - 2√(3)k), so x = 4.

If P = 3i + √(3)j + 2k and Q = 4i + √(3)j + 2.5k then, the unit… | JEE Main 2023 (25 Jan, Shift 1)

Q8JEE Main 2023 · 25 JAN · SHIFT 1Medium

Vector (Cross) Product

If $\vec{P} = 3\hat{i} + \sqrt{3}\hat{j} + 2\hat{k}$ and $\vec{Q} = 4\hat{i} + \sqrt{3}\hat{j} + 2.5\hat{k}$ then, the unit vector in the direction of $\vec{P} \times \vec{Q}$ is $\frac{1}{x}\left(\sqrt{3}\hat{i} + \hat{j} - 2\sqrt{3}\hat{k}\right)$ . The value of $x$ is _________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 4

$\vec{P} \times \vec{Q} = \hat{i}(2.5\sqrt{3} - 2\sqrt{3}) - \hat{j}(7.5 - 8) + \hat{k}(3\sqrt{3} - 4\sqrt{3})$
$= \frac{1}{2}\left(\sqrt{3}\hat{i} + \hat{j} - 2\sqrt{3}\hat{k}\right)$
Magnitude of $\sqrt{3}\hat{i} + \hat{j} - 2\sqrt{3}\hat{k}$ is $\sqrt{3 + 1 + 12} = 4$ .
Unit vector $= \frac{1}{4}\left(\sqrt{3}\hat{i} + \hat{j} - 2\sqrt{3}\hat{k}\right)$ , so $x = 4$ .

Report a mistake

Vector (Cross) Product question from JEE Main 2023 (25 Jan, Shift 1)

More questions on Vector (Cross) Product