Vardaan Institute - Detailed Solutions: Master Sheet (Vectors)

Section A: Vector Addition & Components Solutions

Two vectors $\vec{A}$ and $\vec{B}$ of magnitude 6 units and 8 units act at an angle of $60^\circ$. Find magnitude of resultant vector $\vec{R}$.

Sol: $R = \sqrt{A^2 + B^2 + 2AB\cos 60^\circ} = \sqrt{6^2 + 8^2 + 2(6)(8)(0.5)} = \sqrt{36 + 64 + 48} = \sqrt{148} \approx 12.17\text{ units}$.

Find a unit vector parallel to the resultant of $\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{B} = \hat{i} - 2\hat{j} + 4\hat{k}$.

Sol: $\vec{R} = \vec{A} + \vec{B} = 3\hat{i} + \hat{j} + 3\hat{k}$. $|\vec{R}| = \sqrt{3^2 + 1^2 + 3^2} = \sqrt{19}$. Unit vector $\hat{r} = \frac{3\hat{i} + \hat{j} + 3\hat{k}}{\sqrt{19}}$.

Section B: Dot Product & Projections Solutions

Find the angle between $\vec{A} = 3\hat{i} + 4\hat{j}$ and $\vec{B} = 4\hat{i} - 3\hat{j}$.

Sol: $\vec{A} \cdot \vec{B} = (3)(4) + (4)(-3) = 12 - 12 = 0 \implies \theta = 90^\circ$ (perpendicular).

Find the value of $m$ for which vectors $\vec{A} = 2\hat{i} + m\hat{j} + \hat{k}$ and $\vec{B} = 4\hat{i} - 2\hat{j} - 2\hat{k}$ are perpendicular.

Sol: $\vec{A} \cdot \vec{B} = 0 \implies 2(4) + m(-2) + 1(-2) = 0 \implies 8 - 2m - 2 = 0 \implies 2m = 6 \implies m = 3$.

Section C: Cross Product & Applications Solutions

Find area of the parallelogram determined by vectors $\vec{A} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{B} = 2\hat{i} + \hat{j} - 4\hat{k}$.

Sol: $\vec{A} \times \vec{B} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-2&3\\2&1&-4\end{vmatrix} = 5\hat{i} + 10\hat{j} + 5\hat{k}$. Area $= |\vec{A}\times\vec{B}| = \sqrt{25 + 100 + 25} = \sqrt{150} = 5\sqrt{6}\text{ sq units}$.

Calculate torque $\vec{\tau} = \vec{r} \times \vec{F}$ where $\vec{r} = 7\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{F} = -3\hat{i} + \hat{j} + 5\hat{k}$.

Sol: $\vec{\tau} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\7&3&1\\-3&1&5\end{vmatrix} = \hat{i}(15-1) - \hat{j}(35+3) + \hat{k}(7+9) = 14\hat{i} - 38\hat{j} + 16\hat{k}\text{ N}\cdot\text{m}$.

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