If \overrightarrow P = 3\widehat i + \sqrt 3 \widehat j + 2\widehat k and \overrightarrow Q = 4\widehat i + \sqrt 3 \widehat j + 2.5\widehat k then, the unit vector in the direction of \overrightarrow P \times \overrightarrow Q is 1 \over x( \sqrt 3 \widehat i + \widehat j - 2\sqrt 3 \widehat k ) . The value of x is \underline\hspace2cm .

Vector Algebra PYQ — JEE Main 2023 · 25 Jan · Shift 1 · Q63 with Solution

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Correct answer: 4

We are given $\vec P = 3\hat i + \sqrt{3}\hat j + 2\hat k,$ $\vec Q = 4\hat i + \sqrt{3}\hat j + 2.5\hat k = 4\hat i + \sqrt{3}\hat j + \frac{5}{2}\hat k.$

We need the unit vector along $\vec P \times \vec Q$ .

\begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & \sqrt{3} & 2 \\ 4 & \sqrt{3} & \frac{5}{2} \end{vmatrix}.$$ Expanding, $$\vec P \times \vec Q = \hat i\left(\sqrt{3}\cdot \frac{5}{2} - 2\sqrt{3}\right) - \hat j\left(3\cdot \frac{5}{2} - 2\cdot 4\right) + \hat k\left(3\sqrt{3} - 4\sqrt{3}\right).$$ Now simplify each component: - $\hat i$ component: $$\sqrt{3}\left(\frac{5}{2} - 2\right)=\sqrt{3}\cdot \frac{1}{2}=\frac{\sqrt{3}}{2}.$$ - $\hat j$ component: $$-\left(\frac{15}{2}-8\right)=-\left(\frac{15-16}{2}\right)= -\left(-\frac{1}{2}\right)=\frac{1}{2}.$$ - $\hat k$ component: $$3\sqrt{3}-4\sqrt{3}=-\sqrt{3}.$$ So, $$\vec P \times \vec Q = \frac{\sqrt{3}}{2}\hat i + \frac{1}{2}\hat j - \sqrt{3}\hat k.$$ 3. Factor out $\frac{1}{2}$: $$\vec P \times \vec Q = \frac{1}{2}\left(\sqrt{3}\hat i + \hat j - 2\sqrt{3}\hat k\right).$$ This matches the given direction vector. 4. Find its magnitude: $$\left|\sqrt{3}\hat i + \hat j - 2\sqrt{3}\hat k\right| = \sqrt{(\sqrt{3})^2 + 1^2 + (-2\sqrt{3})^2}$$ $$= \sqrt{3 + 1 + 12} = \sqrt{16} = 4.$$ Hence the unit vector in that direction is $$\frac{1}{4}\left(\sqrt{3}\hat i + \hat j - 2\sqrt{3}\hat k\right).$$ Comparing with $$\frac{1}{x}\left(\sqrt{3}\hat i + \hat j - 2\sqrt{3}\hat k\right),$$ we get $$x=4.$$

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Correct answer: 4

We are given $\vec P = 3\hat i + \sqrt{3}\hat j + 2\hat k,$ $\vec Q = 4\hat i + \sqrt{3}\hat j + 2.5\hat k = 4\hat i + \sqrt{3}\hat j + \frac{5}{2}\hat k.$

We need the unit vector along $\vec P \times \vec Q$ .

\begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & \sqrt{3} & 2 \\ 4 & \sqrt{3} & \frac{5}{2} \end{vmatrix}.$$ Expanding, $$\vec P \times \vec Q = \hat i\left(\sqrt{3}\cdot \frac{5}{2} - 2\sqrt{3}\right) - \hat j\left(3\cdot \frac{5}{2} - 2\cdot 4\right) + \hat k\left(3\sqrt{3} - 4\sqrt{3}\right).$$ Now simplify each component: - $\hat i$ component: $$\sqrt{3}\left(\frac{5}{2} - 2\right)=\sqrt{3}\cdot \frac{1}{2}=\frac{\sqrt{3}}{2}.$$ - $\hat j$ component: $$-\left(\frac{15}{2}-8\right)=-\left(\frac{15-16}{2}\right)= -\left(-\frac{1}{2}\right)=\frac{1}{2}.$$ - $\hat k$ component: $$3\sqrt{3}-4\sqrt{3}=-\sqrt{3}.$$ So, $$\vec P \times \vec Q = \frac{\sqrt{3}}{2}\hat i + \frac{1}{2}\hat j - \sqrt{3}\hat k.$$ 3. Factor out $\frac{1}{2}$: $$\vec P \times \vec Q = \frac{1}{2}\left(\sqrt{3}\hat i + \hat j - 2\sqrt{3}\hat k\right).$$ This matches the given direction vector. 4. Find its magnitude: $$\left|\sqrt{3}\hat i + \hat j - 2\sqrt{3}\hat k\right| = \sqrt{(\sqrt{3})^2 + 1^2 + (-2\sqrt{3})^2}$$ $$= \sqrt{3 + 1 + 12} = \sqrt{16} = 4.$$ Hence the unit vector in that direction is $$\frac{1}{4}\left(\sqrt{3}\hat i + \hat j - 2\sqrt{3}\hat k\right).$$ Comparing with $$\frac{1}{x}\left(\sqrt{3}\hat i + \hat j - 2\sqrt{3}\hat k\right),$$ we get $$x=4.$$