Triangle Law of Vector Addition: Statement & Derivation

Ans. Vector quantities are those quantities that have both magnitude and direction. 

Ans. The two laws of vector addition are:

• Triangle law of vector addition
• Parallelogram law of vector addition

Ans. Since the vector is to be directed from P to Q, clearly P is the initial point

and Q is the terminal point. So, the required vector joining P and Q is the vector PQ ,

\(\bar{PQ}\) = ( – 1 – 2)\(\hat{i}\) + ( – 2 – 3)\(\hat{j}\) + ( – 4 – 0)\(\hat{k}\)

\(\bar{PQ}\) = – 3\(\hat{i}\) – 5\(\hat{j}\) – 4\(\hat{k}\)

Ans. We have \(3 \hat{i} + 2 \hat{j} + 9 \hat{k}\) and \(\hat{i} - 2p \hat{j} + 3 \hat{k}\) which are parallel vectors, so their directional ratios will be proportional. 

∴ \(\frac{3}{1} = \frac{2}{-2p} = \frac{9}{3}\)⇒\( \frac{2}{-2p} = \frac{3}{1}\)

⇒ – 6p = 2 ⇒ p = \(\frac{2}{-6}\)⇒ p = – \(\frac{1}{3}\)

Ans. It is given, \(\overrightarrow{a}\) = \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)

Now, unit vector in the direction of \(\overrightarrow{a}\) is 

\(\hat{a} \) = \(\frac{\overrightarrow{a}}{ |\overrightarrow{a}|}\) = \(\frac{ \hat{i} + \hat{j} + \hat{k}}{ \sqrt{(1)^2 + (1)^2 + (1)^2}}\)

= \(\frac{ \hat{i} + \hat{j} + \hat{k}}{ \sqrt{3}}\) = \(\frac{1}{\sqrt{3}}\)\(\hat{i}\) + \(\frac{1}{\sqrt{3}}\)\(\hat{j}\) + \(\frac{1}{\sqrt{3}}\)\(\hat{k}\)

Therefore, the cosine of angle that the given vector makes with the Y-axis is 1/√3. 

Ans. Firstly, find the resultant of the vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\), \(\overrightarrow{a} + \overrightarrow{b} \). Then we have to find the unit vector in the direction of \(\overrightarrow{a} + \overrightarrow{b} \) i.e., the unit vector is multiplied by 5.

Given, \(\overrightarrow{a} = 2\hat{i} + 3\hat{j} - \hat{k}\) and \(\overrightarrow{b} = \hat{i} - 2\hat{j} + \hat{k}\)

Now, resultant of the above vectors = \(\overrightarrow{a} + \overrightarrow{b} \)

= \(( 2\hat{i} + 3\hat{j} - \hat{k})\) + \((\hat{i} - 2\hat{j} + \hat{k})\) = 3\(\hat{i} + \hat{j}\) (1)

Let, \(\overrightarrow{a} + \overrightarrow{b} \) = \(\overrightarrow{c}\)

∴ \(\overrightarrow{c}\) = 3\(\hat{i} + \hat{j}\)

now, unit vector \(\hat{c}\) in the direction of \(\overrightarrow{c}\), i.e.

= \(\frac{\overrightarrow{c}}{ |\overrightarrow{c}|}\)                                           (1)

= \(\frac{3 \hat{i} + \hat{j}}{\sqrt{(3)^2 + (1)^2}}\) = \(\frac{3\hat{i} + \hat{j}}{\sqrt{10}}\)= \(\frac{3}{\sqrt{10}}\hat{i} + \frac{1}{\sqrt{10}}\hat{j} \) (1)

Hence, vector of magnitude 5 units and parallel to resultant of \(\overrightarrow{a}\) and \(\overrightarrow{b}\) is

5(\(\frac{3}{\sqrt{10}}\hat{i} + \frac{1}{\sqrt{10}}\hat{j} \)) or \(\frac{15}{\sqrt{10}}\hat{i} + \frac{5}{\sqrt{10}}\hat{j} \) (1)

Ans. Firstly, we have to find the vector \(2\overrightarrow{a} - \overrightarrow{b} + 3\overrightarrow{c}\), then find the vector in the direction of \(2\overrightarrow{a} - \overrightarrow{b} + 3\overrightarrow{c}\), i.e., the unit vector multiplied by 6.

Given, ; \(\overrightarrow{a} = \hat{i} + \hat{j} + \hat{k}, \overrightarrow{b} = 4\hat{i} - 2\hat{j} + 3\hat{k}\)

and \(\overrightarrow{c} = \hat{i} - 2\hat{j} + \hat{k}\)

Therefore, \(2\overrightarrow{a} - \overrightarrow{b} + 3\overrightarrow{c}\)

= 2(\(\hat{i} + \hat{j} + \hat{k}\)) – (\(4\hat{i} - 2\hat{j} + 3\hat{k}\)) + 3(\(\hat{i} - 2\hat{j} + \hat{k}\))

= 2\(\hat{i}\) + 2\(\hat{j}\) + 2\(\hat{k}\) – 4\(\hat{i}\) + 2\(\hat{j}\) – 3\(\hat{k}\) + 3\(\hat{i}\) – 6\(\hat{j}\) + 3\(\hat{k}\)

⇒ \(2\overrightarrow{a} - \overrightarrow{b} + 3\overrightarrow{c}\) = \(\hat{i} - 2\hat{j} + 2\hat{k}\) (1½)

Now, a unit vector in the direction of vector

\(2\overrightarrow{a} - \overrightarrow{b} + 3\overrightarrow{c}\) = \(\frac{2\overrightarrow{a} - \overrightarrow{b} + 3\overrightarrow{c}}{|2\overrightarrow{a} - \overrightarrow{b} + 3\overrightarrow{c}|}\)

= \(\frac{\hat{i} - 2\hat{j} + 2\hat{k}}{\sqrt{(1)^2 + (-2)^2 + (2)^2}}\) = \(\frac{\hat{i} - 2\hat{j} + 2\hat{k}}{\sqrt{9}}\)

= \(\frac{\hat{i} - 2\hat{j} + 2\hat{k}}{3}\) = \(\frac{1}{3}\)\(\hat{i}\) – \(\frac{2}{3}\)\(\hat{j}\) + \(\frac{2}{3}\)\(\hat{k}\) (1½)

Hence, vector of magnitude 6 units parallel to the vector \(2\overrightarrow{a} - \overrightarrow{b} + 3\overrightarrow{c}\) = 6(\(\frac{1}{3}\)\(\hat{i}\) – \(\frac{2}{3}\)\(\hat{j}\) + \(\frac{2}{3}\)\(\hat{k}\))

= \(2\hat{i} - 4\hat{j} + 4\hat{k}\) (1)

Ans. Given, \(\overrightarrow{OP}\) = position vector of P = \(\overrightarrow{OQ}\)

and = \(\overrightarrow{OR}\) position vector of Q = \(\overrightarrow{a} - 3\overrightarrow{b} \)

Let \(\overrightarrow{OR}\) be the position vector of the point R which, divides PQ in the ratio 1: 2 externally

Now, it must be shown that P is the midpoint of RQ.

i.e. \(\overrightarrow{OP}\) = \(\frac{\overrightarrow{OR} + \overrightarrow{OQ}}{2}\)

we have, \(\overrightarrow{OR}\) = \(3 \overrightarrow{a} + 5 \overrightarrow{b}\), \(\overrightarrow{OQ}\) = \(\overrightarrow{a} - 3 \overrightarrow{b}\)

∴ \(\frac{\overrightarrow{OR} + \overrightarrow{OQ}}{2}\) = \(\frac{(3 \overrightarrow{a} + 5 \overrightarrow{b}) + ( \overrightarrow{a} - 3\overrightarrow{b})}{2}\)

= \(\frac{4 \overrightarrow{a} + 2 \overrightarrow{b}}{2}\)

= \(\frac{ 2 (2\overrightarrow{a} + \overrightarrow{b})}{2}\) = \(2\overrightarrow{a} + \overrightarrow{b}\)

= \(\overrightarrow{OP}\) [ \(\because\) \(\overrightarrow{OP}\) = \(2\overrightarrow{a} + \overrightarrow{b}\), given] (1½)

Therefore, P is the mid-point of the line segment R.

Ans. In order to find a vector in the direction of a given vector, first of all we find the unit vector in the direction of the direct vector and then multiply it with the given magnitude.

Let \(\overrightarrow{a}\) = 2\(\hat{i}\) – 3\(\hat{j}\) + 6\(\hat{k}\)

|\(\overrightarrow{a}\)| = \(\sqrt{(2)^2 + (-3)^2 + (6)^2}\)

= \(\sqrt{4+9+36}\)

= \(\sqrt{49}\) = 7 units

The unit vector in the direction of the given vector \(\overrightarrow{a}\) is

 \(\hat{a} \) = \(\frac{\overrightarrow{a}}{ |\overrightarrow{a}|}\) = \(\frac{1}{7}\) (2\(\hat{i}\) – 3\(\hat{j}\) + 6\(\hat{k}\))

= \(\frac{2}{7}\)\(\hat{i}\) – \(\frac{3}{7}\)\(\hat{j}\) + \(\frac{6}{7}\)\(\hat{k}\)

Therefore the vector of magnitude equal to 21 units and in the direction of \(\overrightarrow{a}\) is,

21\(\hat{a} \) = 21(\(\frac{2}{7}\)\(\hat{i}\) – \(\frac{3}{7}\)\(\hat{j}\) + \(\frac{6}{7}\)\(\hat{k}\))

= 6\(\hat{i}\) – 9\(\hat{j}\)+ 18\(\hat{k}\)