Addition Of Vectors (VSAQs)
Maths-1A | 4. Addition of Vectors – VSAQs:
Welcome to VSAQs in Chapter 4: Addition Of Vectors. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.
VSAQ-1 : Find the unit vector in the direction of vector ¯a = 2¯i + 3¯j + ¯k
$$\text{Given } \overline{a} = 2\overline{i} + 3\overline{j} + \overline{k} \text{ then}$$
$$|\overline{a}| = \sqrt{2^2 + 3^2 + 1^2}$$
$$= \sqrt{4 + 9 + 1}$$
$$= \sqrt{14}$$
$$\text{Unit vector} = \frac{\overline{a}}{|\overline{a}|} = \frac{2\overline{i} + 3\overline{j} + \overline{k}}{\sqrt{14}}$$
VSAQ-2 : Find a vector in the direction of vector ¯a = ¯i – 2¯j that has magnitude 7 units
$$\text{Given } \overline{a} = \overline{i} – 2\overline{j} \text{ then}$$
$$|\overline{a}| = \sqrt{1^2 + (-2)^2}$$
$$= \sqrt{1 + 4}$$
$$= \sqrt{5}$$
$$\text{Unit vector} = \frac{\overline{a}}{|\overline{a}|} = \frac{\overline{i} – 2\overline{j}}{\sqrt{5}}$$
$$\text{Vector of magnitude 7} = 7 \left( \frac{\overline{i} – 2\overline{j}}{\sqrt{5}} \right)$$
VSAQ-3 : Let ¯a = ¯i + 2¯j + 3¯k and ¯b = 3¯i + ¯j. Find a unit vector in the direction of ¯a + ¯b
$$\text{Given } \overline{a} = \overline{i} + 2\overline{j} + 3\overline{k} \text{ and } \overline{b} = 3\overline{i} + \overline{j} \text{ then}$$
$$\overline{a} + \overline{b} = (\overline{i} + 2\overline{j} + 3\overline{k}) + (3\overline{i} + \overline{j})$$
$$= 4\overline{i} + 3\overline{j} + 3\overline{k}$$
$$|\overline{a} + \overline{b}| = \sqrt{4^2 + 3^2 + 3^2}$$
$$= \sqrt{16 + 9 + 9}$$
$$= \sqrt{34}$$
$$\text{Required unit vector} = \frac{\overline{a} + \overline{b}}{|\overline{a} + \overline{b}|}$$
$$= \frac{4\overline{i} + 3\overline{j} + 3\overline{k}}{\sqrt{34}}$$
VSAQ-4 : Let ¯a = 2¯i + 4¯j – 5¯k, ¯b = ¯i + ¯j + ¯k, ¯c = ¯j + 2¯k. Find the unit vector in the opposite direction of ¯a + ¯b + ¯c
$$\text{Given } \overline{a} = 2\overline{i} + 4\overline{j} – 5\overline{k}$$
$$\overline{b} = \overline{i} + \overline{j} + \overline{k}$$
$$\overline{c} = 0\overline{i} + \overline{j} + 2\overline{k} \text{ then}$$
$$\overline{a} + \overline{b} + \overline{c}$$
$$= (2\overline{i} + 4\overline{j} – 5\overline{k}) + (\overline{i} + \overline{j} + \overline{k}) + (0\overline{i} + \overline{j} + 2\overline{k})$$
$$= 3\overline{i} + 6\overline{j} – 2\overline{k}$$
$$|\overline{a} + \overline{b} + \overline{c}| = \sqrt{3^2 + 6^2 + (-2)^2}$$
$$= \sqrt{9 + 36 + 4}$$
$$= \sqrt{49}$$
$$= 7$$
$$\text{Opposite unit vector} = -\frac{\overline{a} + \overline{b} + \overline{c}}{|\overline{a} + \overline{b} + \overline{c}|}$$
$$= -\frac{3\overline{i} + 6\overline{j} – 2\overline{k}}{7}$$
VSAQ-5 : If ¯a = 2¯i + 5¯j + ¯k, ¯b = 4¯i + m¯j + n¯k are collinear vectors then find m,n
$$\text{Given vectors } \overline{a} = 2\overline{i} + 5\overline{j} + \overline{k}$$
$$\overline{b} = 4\overline{i} + m\overline{j} + n\overline{k} \text{ are collinear}$$
$$\frac{2}{4} = \frac{5}{m} = \frac{1}{n}$$
$$\frac{1}{2} = \frac{5}{m}$$
$$m = 2 \times 5 = 10$$
$$\frac{1}{n} = \frac{1}{2}$$
$$n = 2$$
$$m = 10, \quad n = 2$$
VSAQ-6 : If -3¯i + 4¯j + λ¯k, μ¯i + 8¯j + 6¯k are collinear vectors then find λ & μ
$$\text{Given that the vectors } \overline{a} = -3\overline{i} + 4\overline{j} + \lambda\overline{k}$$
$$\overline{b} = \mu\overline{i} + 8\overline{j} + 6\overline{k} \text{ are collinear}$$
$$\frac{-3}{\mu} = \frac{4}{8} = \frac{\lambda}{6}$$
$$\frac{-3}{\mu} = \frac{1}{2}$$
$$\mu = 2 \times -3 = -6$$
$$\frac{\lambda}{6} = \frac{1}{2}$$
$$\lambda = 6 \times \frac{1}{2} = 3$$
$$\lambda = 3, \quad \mu = -6$$
VSAQ-7 : Find the vector equation of the line passing through the point 2¯i + ¯j + 3¯k and parallel to the vector 4¯i – 2¯j + 3¯k
$$\text{Given point } A(\overline{a}) = 2\overline{i} + \overline{j} + 3\overline{k} \text{ and}$$
$$\text{parallel vector } \overline{b} = 4\overline{i} – 2\overline{j} + 3\overline{k}$$
$$\text{Vector equation of the line is}$$
$$\overline{r} = \overline{a} + t\overline{b}, \quad t \in \mathbb{R}$$
$$\overline{r} = (2\overline{i} + \overline{j} + 3\overline{k}) + t(4\overline{i} – 2\overline{j} + 3\overline{k}), \quad t \in \mathbb{R}$$
VSAQ-8 : Find the vector equation of the line passing through the points 2¯i + ¯j + 3¯k, -4¯i + 3¯j – ¯k
$$\text{Given points } A(\overline{a}) = 2\overline{i} + \overline{j} + 3\overline{k}$$
$$B(\overline{b}) = -4\overline{i} + 3\overline{j} – \overline{k}$$
$$\text{Vector equation of the line is}$$
$$\overline{r} = (1 – t)\overline{a} + t\overline{b}, \quad t \in \mathbb{R}$$
$$\overline{r} = (1 – t)(2\overline{i} + \overline{j} + 3\overline{k}) + t(-4\overline{i} + 3\overline{j} – \overline{k}), \quad t \in \mathbb{R}$$
VSAQ-9 : Find the vector equation of the plane passing through the points ¯i – 2¯j + 5¯k, -5¯j – ¯k, -3¯i + 5¯j
$$\text{Given } A(\overline{a}) = \overline{i} – 2\overline{j} + 5\overline{k},$$
$$B(\overline{b}) = -5\overline{j} – \overline{k},$$
$$C(\overline{c}) = -3\overline{i} + 5\overline{j}$$
$$\text{Vector equation of the plane is}$$
$$\overline{r} = (1 – s – t)\overline{a} + s\overline{b} + t\overline{c}, \quad s, t \in \mathbb{R}$$
$$\overline{r} = (1 – s – t)(\overline{i} – 2\overline{j} + 5\overline{k}) + s(-5\overline{j} – \overline{k}) + t(-3\overline{i} + 5\overline{j}), \quad s, t \in \mathbb{R}$$
VSAQ-10 : Find the vector equation of the plane passing through the points (0, 0, 0), (0, 5, 0) and (2, 0, 1)
$$\text{Given } A(\overline{a}) = \overline{0},$$
$$B(\overline{b}) = 5\overline{j},$$
$$C(\overline{c}) = 2\overline{i} + \overline{k}$$
$$\text{Vector equation of the plane is}$$
$$\overline{r} = (1 – s – t)\overline{a} + s\overline{b} + t\overline{c}, \quad s, t \in \mathbb{R}$$
$$\overline{r} = (1 – s – t)\overline{0} + s(5\overline{j}) + t(2\overline{i} + \overline{k})$$
$$\overline{r} = (5\overline{j})s + t(2\overline{i} + \overline{k}), \quad s, t \in \mathbb{R}$$