Question

The components of a vector A=2i-3j along the vector i+j ?
a)5
b)5/√2
c) 10√2
d) -1/√2​

Answer 1

Given:

$\rm \overrightarrow{A} = 2 \hat{i} - 3 \hat{j}$

$\rm \overrightarrow{B} = \hat{i} + \hat{j}$

To Find:

Component of $\sf \overrightarrow{A}$ along $\sf \overrightarrow{B}$

Answer:

Magnitude of $\sf \overrightarrow{A}$ & $\sf \overrightarrow{B}$:

$\rm | \overrightarrow{A} | = \sqrt{ {2}^{2} + {3}^{2} } \\ \\ \rm A = \sqrt{4 + 9} \\ \\ \rm A = \sqrt{13} \\ \\ \\ \rm | \overrightarrow{B} | = \sqrt{ {1}^{2} + {1}^{2} } \\ \\ \rm B = \sqrt{1 + 1} \\ \\ \rm B = \sqrt{2}$

Dot product of $\sf \overrightarrow{A}$ & $\sf \overrightarrow{B}$:

$\rm \overrightarrow{A}. \overrightarrow{B} = (2 \hat{i} - 3 \hat{j})( \hat{i} + \hat{j}) \\ \\ \rm \overrightarrow{A}. \overrightarrow{B} =2 - 3 \\ \\ \rm \overrightarrow{A}. \overrightarrow{B} = - 1$

Suppose θ is the angle between the vectors.

The component of $\sf \overrightarrow{A}$ along $\sf \overrightarrow{B}$ is A cosθ.

Thus,

$\rm \implies \overrightarrow{A}. \overrightarrow{B} =AB \: cos \theta \\ \\ \rm \implies A \: cos \theta = \dfrac{ \overrightarrow{A}. \overrightarrow{B}}{B} \\ \\ \rm \implies A \: cos \theta = - \dfrac{1}{ \sqrt{2} }$

$\therefore$ Component of $\sf \overrightarrow{A}$ along $\sf \overrightarrow{B}$ = $\rm - \dfrac{1}{ \sqrt{2} }$

Correct Option: $\boxed{\mathfrak{(d) \ - \dfrac{1}{ \sqrt{2} } }}$

Answer 2

Acosθ = vecA.vecB/B

= (2i-3j)(i+j)/√1²+1²

= 2-3/√1+1

=-1/√2