4.1 Resolution of a Vector:

2.1 Concept of Unit Vector

3.1 Parallelogram Law of Vector Addition
3.2 Properties of Vector Addition

4.1 Resolution of a Vector:

8.1 Scalar (or dot) products of two vectors
8.2 Vector or Cross Product of two Vectors

10.1 Collinearity of Vectors
10.2 Collinearity of points

The splitting of a vector into its components is called resolution of a vector. Generally there are three components but in this chapter we are assuming $2-D$ plane, so consider only two components. Component along $x$-axis is called $x$-component and component along $y$-axis is called $y$-component.

Consider a vector $\overrightarrow P $ acting at a point making an angle $\theta $ with the positive $x$-axis. Vector $\overrightarrow P $ is represented by a line $OA$. Now from point $A$, draw a perpendicular $AB$ on $x$-axis. If $OB$ and $BA$ represents two vectors. Vector $OB$ is parallel to $x$-axis and vector $BA$ is parallel to $y$-axis. Magnitude of these vectors are given as ${P_x}$ and ${P_y}$ where ${P_x}$ - horizontal component of $\overrightarrow P $ and ${P_y}$ - vertical component of $\overrightarrow P $.

In a right angle triangle $OAB$, magnitude of $x$-component ${P_x}$ is given as $$\begin{equation} \begin{aligned} cos\theta = \frac{{\overrightarrow {OB} }}{{\overrightarrow {OA} }} \\ \overrightarrow {OB} = \overrightarrow {OA} cos\theta \\ {P_x} = Pcos\theta \\\end{aligned} \end{equation} $$

and magnitude of $y$-component ${P_y}$ is given as $$\begin{equation} \begin{aligned} sin\theta = \frac{{\overrightarrow {AB} }}{{\overrightarrow {OA} }} \\ \overrightarrow {AB} = \overrightarrow {OA} sin\theta \\ {P_y} = Psin\theta \\\end{aligned} \end{equation} $$

If $a$ and $b$ are any two vectors given in the component form $$\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$$ and $$\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k$$ Then

(i) the sum (or resultant) of the vectors $\overrightarrow a $ and $\overrightarrow b $ is given as $$({a_1} + {b_1})\widehat i + ({a_2} + {b_2})\widehat j + ({a_3} + {b_3})\widehat k$$

(ii) the difference of the $\overrightarrow a $ and $\overrightarrow b $ i.e., $\overrightarrow a - \overrightarrow b $ is given as

$$({a_1} - {b_1})\widehat i + ({a_2} - {b_2})\widehat j + ({a_3} - {b_3})\widehat k$$

(iii) the vectors $\overrightarrow a $ and $\overrightarrow b $ are equal if and only if $${a_1} = {b_1},\ {a_2} = {b_2},\ {a_3} = {b_3}$$

(iv) the multiplication of vector $\overrightarrow a $ by any scalar $\lambda$ is given by $$\lambda \vec a = \lambda \vec a = (\lambda {a_1})\hat i + (\lambda {a_2})\hat j + (\lambda {a_3})\hat k$$

We observe that whatever be the value of $\lambda $, the vector $\lambda \overrightarrow a $ is always collinear to the $\overrightarrow a $ are collinear if and only if there exists a nonzero scalar $\lambda $ such that $\overrightarrow b = \lambda \overrightarrow a $.

If the vectors $\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$ and $\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k$ then the two vectors are said to be collinear if and only if

$$\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k = \lambda ({a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k)$$

$$\begin{equation} \begin{aligned} {b_1} = \lambda {a_1} \\ {b_2} = \lambda {a_2} \\ {b_3} = \lambda {a_3} \\\end{aligned} \end{equation} $$

$$\frac{{{b_1}}}{{{a_1}}} = \frac{{{b_2}}}{{{a_2}}} = \frac{{{b_3}}}{{{a_3}}} = \lambda $$

(v) $\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$ then , ${a_1},{a_2},{a_3}$ are the direction ratios of $\overrightarrow a $.

(vi) If $l,\ m\ and\ n$ are the direction cosines of a vector, then $$l\widehat i + m\widehat j + n\widehat k = (\cos \alpha )\widehat i + (\cos \beta )\widehat j + (\cos \lambda )\widehat k$$ is the unit vector in the direction of the vector where $\alpha,\beta,\lambda $ are angles which the vector makes with $x$, $y$ and $z$ axes respectively.

Question 3. Find the values of $x$, $y$ and $z$ for which $\overrightarrow a = 2\widehat i + y\widehat j + 6\widehat k$ and $\overrightarrow b = x\widehat i + 3\widehat j + z\widehat k$ are equal.

Solution: Given vectors are equal if the corresponding components are equal so $x = 2$, $y = 3$ and $z = 6$.

Question 4. Find the values of $x$ and $y$ for which vectors $\overrightarrow a = (2 - x)\widehat i + (y + 3)\widehat j + 6\widehat k$ and $\overrightarrow b = 3\widehat i + 3\widehat j + 7\widehat k$ are equal.

Solution: $$\begin{equation} \begin{aligned} 2 - x = 3 \\ or,x = 2 - 3 \\ x = - 1, \\ y + 3 = 3 \\ y = 0 \\\end{aligned} \end{equation} $$

Question 5. $\overrightarrow a = 7\widehat j + 3\widehat k$ and $\overrightarrow b = 3\widehat j + 7\widehat k$. Are the vectors equal?

$$\begin{equation} \begin{aligned} \overrightarrow b = 3\widehat j + 7\widehat k \\ \left| {\overrightarrow b } \right| = \sqrt {{3^2} + {7^2}} = \sqrt {58} \\\end{aligned} \end{equation} $$

$$\begin{equation} \begin{aligned} \overrightarrow a = 7\widehat j + 3\widehat k \\ \left| {\overrightarrow a } \right| = \sqrt {{7^2} + {3^2}} = \sqrt {58} \\\end{aligned} \end{equation} $$

We can conclude that the magnitudes are equal but direction are different so vectors are not equal.

Question 6. Find the unit vector in the direction of the given vector $\overrightarrow a = 7\widehat i + 3\widehat j + \sqrt 6 \widehat k$.

Solution: The unit vector in the direction of vector $\overrightarrow a = 7\widehat i + 3\widehat j + \sqrt 6 \widehat k$ given by

$$\begin{equation} \begin{aligned} \widehat a = \frac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}} \\ \widehat a = \frac{{7\widehat i + 3\widehat j + \sqrt 6 \widehat k}}{{\sqrt {{7^2} + {3^2} + {{(\sqrt 6 )}^2}} }} \\ \widehat a = \frac{{7\widehat i + 3\widehat j + \sqrt 6 \widehat k}}{{\sqrt {49 + 9 + 6} }} \\ \widehat a = \frac{{7\widehat i + 3\widehat j + \sqrt 6 \widehat k}}{8} \\\end{aligned} \end{equation} $$

Question 7. Find a vector in the direction of vector $\overrightarrow a = 2\widehat i + 3\widehat j + 5\widehat k$ that has a magnitude of 6 units.

Solution: The unit vector in the direction of the given vector is given by

$$\begin{equation} \begin{aligned} \widehat a = \frac{{2\widehat i + 3\widehat j + 5\widehat k}}{{\sqrt {{2^2} + {3^2} + {5^2}} }} \\ \widehat a = \frac{{2\widehat i + 3\widehat j + 5\widehat k}}{{\sqrt {4 + 9 + 25} }} \\ \widehat a = \frac{{2\widehat i + 3\widehat j + 5\widehat k}}{{\sqrt {38} }} \\\end{aligned} \end{equation} $$

So the vector having magnitude $6$ units and in the direction of $\overrightarrow a $ is given by $6\widehat a$ which is equal to $$\begin{equation} \begin{aligned} 6\widehat a = 6\left( {\frac{{2\widehat i + 3\widehat j + 5\widehat k}}{{\sqrt {38} }}} \right) \\ 6\widehat a = \frac{{12\widehat i + 18\widehat j + 30\widehat k}}{{\sqrt {38} }} \\\end{aligned} \end{equation} $$

Question 8. Find the unit vector in the direction of the difference of the vectors $\overrightarrow a = 4\widehat i + 6\widehat j + 8\widehat k$ and $\overrightarrow b = 8\widehat i + 9\widehat j + 10\widehat k$.

Solution : The difference of the given vectors is given as

$$\begin{equation} \begin{aligned} \overrightarrow b - \overrightarrow a = (8\widehat i + 9\widehat j + 10\widehat k) - (4\widehat i - 6\widehat j + 8\widehat k) \\ \overrightarrow b - \overrightarrow a = 4\widehat i + 15\widehat j + 2\widehat k \\\end{aligned} \end{equation} $$

Let $\overrightarrow b - \overrightarrow a = \overrightarrow c $

Magnitude of $\overrightarrow c $ is given as $$\left| {\overrightarrow c } \right| = \sqrt {16 + 225 + 4} = \sqrt {245} $$

The required unit vector is given as

$$\begin{equation} \begin{aligned} \widehat c = \frac{{\overrightarrow c }}{{\left| {\overrightarrow c } \right|}} \\ \widehat c = \frac{{4\widehat i + 15\widehat j + 2\widehat k}}{{\sqrt {245} }} \\\end{aligned} \end{equation} $$

Question 9. Write the direction ratios of the vector $\overrightarrow c = 2\widehat i + \widehat j - \widehat k$ and its direction cosines.

Solution: We know that the direction ratios are the respective components of a vector along $x$, $y$ and $z$ direction so direction ratios are $2$, $1$, and $- 1$. Direction cosines $l$, $m$ and $n$ are given as

$$\begin{equation} \begin{aligned} l = \frac{2}{{\sqrt {{2^2} + {1^2} + {{( - 1)}^2}} }} \\ l = \frac{2}{{\sqrt 6 }} \\ m = \frac{1}{{\sqrt 6 }} \\ n = \frac{{ - 1}}{{\sqrt 6 }} \\\end{aligned} \end{equation} $$

Thus direction cosines are $\left( {\frac{2}{{\sqrt 6 }},\frac{1}{{\sqrt 6 }},\frac{{ - 1}}{{\sqrt 6 }}} \right)$