Vardaan Learning Institute

Ans: $|\vec{a}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.

Ans: $|\vec{b}| = \sqrt{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}} = \sqrt{1} = 1$.

Ans: $\hat{a} = \frac{\hat{i} + 2\hat{j} + 2\hat{k}}{\sqrt{1^2+2^2+2^2}} = \frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$.

Ans: Magnitude of given vector is $\sqrt{3^2 + (-4)^2} = 5$. The unit vector is $\frac{3\hat{i} - 4\hat{j}}{5}$. The required vector is $5 \times \left(\frac{3\hat{i} - 4\hat{j}}{5}\right) = 3\hat{i} - 4\hat{j}$.

Ans: $\vec{a} + \vec{b} = (2+1)\hat{i} + (1-1)\hat{j} + (-1+2)\hat{k} = 3\hat{i} + 0\hat{j} + \hat{k} = 3\hat{i} + \hat{k}$.

Ans: $2\vec{a} - \vec{b} = 2(2\hat{i} + \hat{j} - \hat{k}) - (\hat{i} - \hat{j} + 2\hat{k}) = 4\hat{i} + 2\hat{j} - 2\hat{k} - \hat{i} + \hat{j} - 2\hat{k} = 3\hat{i} + 3\hat{j} - 4\hat{k}$.

Ans: $\vec{PQ} = \text{P.V. of } Q - \text{P.V. of } P = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}$.

Ans: $|\vec{PQ}| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt{3}$.

Ans: Yes. Reason: $\vec{b} = -2(\hat{i} - 2\hat{j} + 3\hat{k}) = -2\vec{a}$. Since $\vec{b} = \lambda\vec{a}$, they are collinear.

Ans: $x = 4, y = 3$. Reason: Compare corresponding components.

Ans: $\vec{m} = \frac{(2+4)}{2}\hat{i} + \frac{(3+1)}{2}\hat{j} + \frac{(4-2)}{2}\hat{k} = 3\hat{i} + 2\hat{j} + \hat{k}$.

Proof: $\hat{a} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}$. $\hat{b} = \frac{2\hat{i}+2\hat{j}}{\sqrt{8}} = \frac{2(\hat{i}+\hat{j})}{2\sqrt{2}} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}$. Since $\hat{a} = \hat{b}$, they have the same direction.

Ans: $7 \times \left(\frac{\hat{i} - 2\hat{j}}{\sqrt{5}}\right) = \frac{7}{\sqrt{5}}\hat{i} - \frac{14}{\sqrt{5}}\hat{j}$.

Ans: Let $\vec{c} = \vec{a} + \vec{b} = \hat{i} + 0\hat{j} + 3\hat{k}$. Unit vector = $\frac{\hat{i} + 3\hat{k}}{\sqrt{1^2+3^2}} = \frac{1}{\sqrt{10}}\hat{i} + \frac{3}{\sqrt{10}}\hat{k}$.

Ans: $\vec{a} + \vec{b} = \hat{i} + 2\hat{j} + \hat{k}$. Magnitude = $\sqrt{1^2+2^2+1^2} = \sqrt{6}$.

Ans: Direction ratios are $1, -2, 3$.

Ans: $|\vec{a}| = 3$. Direction cosines are $\frac{1}{3}, \frac{2}{3}, \frac{2}{3}$.

Ans: Let angle be $\alpha$. $l = m = n = \cos\alpha$. So $3\cos^2\alpha = 1 \Rightarrow \cos\alpha = \pm\frac{1}{\sqrt{3}}$. DCs are $\pm\frac{1}{\sqrt{3}}, \pm\frac{1}{\sqrt{3}}, \pm\frac{1}{\sqrt{3}}$.

Ans: $1, 0, 0$.

Ans: $0, 1, 0$.

Ans: $\vec{r} = \frac{2(-\hat{i}+\hat{j}+\hat{k}) + 1(\hat{i}+2\hat{j}-\hat{k})}{3} = \frac{-\hat{i} + 4\hat{j} + \hat{k}}{3}$.

Ans: $\vec{r} = \frac{2(-\hat{i}+\hat{j}+\hat{k}) - 1(\hat{i}+2\hat{j}-\hat{k})}{2-1} = -3\hat{i} + 3\hat{k}$.

Ans: $\vec{AB} = -2\hat{i} - 4\hat{j} + 4\hat{k}$. $|\vec{AB}| = \sqrt{4+16+16} = 6$. DCs are $-\frac{2}{6}, -\frac{4}{6}, \frac{4}{6} = -\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}$.

Ans: $l^2+m^2+n^2=1 \Rightarrow \frac{1}{4} + \frac{1}{4} + n^2 = 1 \Rightarrow n^2 = \frac{1}{2} \Rightarrow n = \pm\frac{1}{\sqrt{2}}$.

Ans: Yes. Reason: $\cos^2 45^\circ + \cos^2 60^\circ + \cos^2 120^\circ = (\frac{1}{\sqrt{2}})^2 + (\frac{1}{2})^2 + (-\frac{1}{2})^2 = \frac{1}{2} + \frac{1}{4} + \frac{1}{4} = 1$. The sum is 1, so it is possible.

Ans: $\frac{(2+4)\hat{i} + (-1+1)\hat{j} + (3-1)\hat{k}}{2} = 3\hat{i} + 0\hat{j} + \hat{k} = 3\hat{i} + \hat{k}$.

Ans: Direction ratios are $(3-2), (1 - (-1)), (5-3) = 1, 2, 2$.

Ans: Magnitude of DRs is $\sqrt{4+9+36} = 7$. DCs are $\frac{2}{7}, \frac{3}{7}, -\frac{6}{7}$.

Ans: $\vec{r} = 14 \times (\text{DCs}) = 14\left(\frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}\right) = 4\hat{i} + 6\hat{j} - 12\hat{k}$.

Ans: No. Reason: A line must have a definite direction, which requires at least one non-zero component.

Ans: $(2)(1) + (1)(-2) + (-3)(1) = 2 - 2 - 3 = -3$.

Ans: $\vec{a}\cdot\vec{b} = 1 - 1 - 1 = -1$. $|\vec{a}| = \sqrt{3}, |\vec{b}| = \sqrt{3}$. $\cos\theta = \frac{-1}{\sqrt{3}\sqrt{3}} = -\frac{1}{3}$. So $\theta = \cos^{-1}\left(-\frac{1}{3}\right)$.

Ans: Projection = $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{2(1) + 3(2) + 2(1)}{\sqrt{1^2+2^2+1^2}} = \frac{10}{\sqrt{6}} = \frac{5\sqrt{6}}{3}$.

Ans: $\vec{a}\cdot\vec{b} = 0 \Rightarrow 2(1) + \lambda(-2) + 1(3) = 0 \Rightarrow 5 - 2\lambda = 0 \Rightarrow \lambda = \frac{5}{2}$.

Ans: $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = |\vec{a}|^2 - |\vec{b}|^2 = 5^2 - 3^2 = 25 - 9 = 16$.

Ans: $|\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b} = 4 + 9 - 2(4) = 5$. Thus $|\vec{a}-\vec{b}| = \sqrt{5}$.

Ans: $(\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a}) = |\vec{x}|^2 - |\vec{a}|^2 = 8$. Since $|\vec{a}|=1$, $|\vec{x}|^2 - 1 = 8 \Rightarrow |\vec{x}|^2 = 9 \Rightarrow |\vec{x}| = 3$.

Ans: $(\hat{i}\cdot\hat{j} + \hat{i}\cdot\hat{k}) + (\hat{j}\cdot\hat{i} + \hat{j}\cdot\hat{k}) + (\hat{k}\cdot\hat{i} + \hat{k}\cdot\hat{j}) = (0+0) + (0+0) + (0+0) = 0$.

Ans: Projection = $\frac{(\hat{i}-\hat{j}) \cdot (\hat{i}+\hat{j})}{|\hat{i}+\hat{j}|} = \frac{1-1}{\sqrt{2}} = 0$.

Ans: Dot product = $(1)(-2) + (-2)(4) + (5)(2) = -2 - 8 + 10 = 0$. Yes, they are orthogonal.

Ans: $\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \frac{1/2}{(1)(1)} = \frac{1}{2}$.

Ans: $\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b}$

Ans: $|\vec{a}|^2 = 0$, so $\vec{a}$ is the zero vector ($\vec{0}$).

Ans: $6|\vec{a}|^2 + 21(\vec{a}\cdot\vec{b}) - 10(\vec{b}\cdot\vec{a}) - 35|\vec{b}|^2 = 6|\vec{a}|^2 + 11\vec{a}\cdot\vec{b} - 35|\vec{b}|^2$.

Ans: Let $\vec{a} = \hat{i}$ and $\vec{b} = \hat{i}+\hat{j}+\hat{k}$. $\cos\theta = \frac{1(1)+0(1)+0(1)}{(1)(\sqrt{3})} = \frac{1}{\sqrt{3}}$. $\theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$.

Ans: $\vec{a} \times \vec{b} = \hat{i}(-2 - 15) - \hat{j}(-4 - 9) + \hat{k}(10 - 3) = -17\hat{i} + 13\hat{j} + 7\hat{k}$.

Ans: $|\vec{a} \times \vec{b}| = \sqrt{(-17)^2 + 13^2 + 7^2} = \sqrt{289 + 169 + 49} = \sqrt{507}$.

Ans: $\vec{a} \times \vec{b} = \hat{i}(1-0) - \hat{j}(1-0) + \hat{k}(0-1) = \hat{i} - \hat{j} - \hat{k}$. Mag = $\sqrt{3}$. Unit vector = $\pm \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$.

Ans: $\vec{a} \times \vec{b} = \hat{i}(-1+21) - \hat{j}(1-6) + \hat{k}(-7+2) = 20\hat{i} + 5\hat{j} - 5\hat{k}$. Area = $\sqrt{400+25+25} = \sqrt{450} = 15\sqrt{2}$.

Ans: $\vec{AB} = 0\hat{i} + \hat{j} + 2\hat{k}$, $\vec{AC} = \hat{i} + 2\hat{j} + 0\hat{k}$. $\vec{AB} \times \vec{AC} = -4\hat{i} + 2\hat{j} - \hat{k}$. Area = $\frac{1}{2}\sqrt{16+4+1} = \frac{\sqrt{21}}{2}$.

Ans: For parallel vectors, ratios of components must be equal. $\frac{2}{p} = \frac{3}{q} = \frac{-1}{-3} \Rightarrow \frac{1}{3}$. So $p=6$ and $q=9$.

Ans: $(\hat{k} - \hat{j}) + (\hat{i} - \hat{k}) + (\hat{j} - \hat{i}) = \vec{0}$.

Ans: $|\vec{a}\times\vec{b}| = ab\sin\theta = 8 \Rightarrow 10\sin\theta = 8 \Rightarrow \sin\theta = 4/5$. Then $\cos\theta = 3/5$ (or $-3/5$). $\vec{a}\cdot\vec{b} = ab\cos\theta = 10(3/5) = 6$ (or $-6$).

Ans: $\vec{d_1} \times \vec{d_2} = \hat{i}(4-6) - \hat{j}(12 - (-2)) + \hat{k}(-9-1) = -2\hat{i} - 14\hat{j} - 10\hat{k}$. Area = $\frac{1}{2}\sqrt{4 + 196 + 100} = \frac{1}{2}\sqrt{300} = \frac{10\sqrt{3}}{2} = 5\sqrt{3}$.

Ans: $\hat{i}\times\hat{i} + \hat{i}\times\hat{j} - \hat{j}\times\hat{i} - \hat{j}\times\hat{j} = \vec{0} + \hat{k} - (-\hat{k}) - \vec{0} = 2\hat{k}$.

Ans: Since they are non-zero, $\sin\theta = 0$. So the angle $\theta$ is $0^\circ$ or $180^\circ$ (they are parallel).

Ans: $\vec{a} \times \vec{b} = \hat{i}(-12) - \hat{j}(9) + \hat{k}(-8) = -12\hat{i} - 9\hat{j} - 8\hat{k}$.

Proof: $\vec{a}\times\vec{a} + \vec{a}\times\vec{b} - \vec{b}\times\vec{a} - \vec{b}\times\vec{b} = \vec{0} + \vec{a}\times\vec{b} - (-\vec{a}\times\vec{b}) - \vec{0} = 2(\vec{a}\times\vec{b})$.

Ans: $\vec{0}$ (Zero vector, since angle between them is $0$).

Ans: $\hat{k}$ and $-\hat{k}$. (Since z-axis is mutually perpendicular to x and y axes).

Ans: $1(1+1) - (-2)(2-0) + 3(2-0) = 2 + 4 + 6 = 12$.

Ans: STP = $1(-3+7) - 3(6-0) + 1(14-0) = 4 - 18 + 14 = 0$. Yes, they are coplanar.

Ans: $2(4-1) - 3(2+3) + 4(-1-6) = 2(3) - 3(5) + 4(-7) = 6 - 15 - 28 = -37$. Volume = $|-37| = 37$ cubic units.

Ans: STP = $1(-3-2\lambda) - (-1)(-9-2) + 1(3\lambda-1) = 0 \Rightarrow -3-2\lambda - 11 + 3\lambda - 1 = 0 \Rightarrow \lambda - 15 = 0 \Rightarrow \lambda = 15$.

Ans: $1 + 1 + 1 = 3$. (Cyclic permutation preserves the value of STP).

Proof: $\vec{a} \times \vec{b}$ is a vector perpendicular to both $\vec{a}$ and $\vec{b}$. The dot product of two perpendicular vectors is always zero.

Ans: $(\hat{i}\cdot\hat{i})\hat{j} - (\hat{i}\cdot\hat{j})\hat{i} = (1)\hat{j} - (0)\hat{i} = \hat{j}$.

Ans: $\hat{i}\cdot\hat{i} + \hat{j}\cdot(-\hat{j}) + \hat{k}\cdot\hat{k} = 1 - 1 + 1 = 1$.

Ans: $(2\times 3\times 4)[\vec{a} \, \vec{b} \, \vec{c}] = 24(5) = 120$.

Ans: $2[\vec{a} \, \vec{b} \, \vec{c}]$. (Standard expansion using distributive property of scalar triple product).