Obtain the component of vector A=2i+3j in the direction of vector -i-j
Answers
Explanation:
Component of any vector:
If \vec{a}a be any vector and we are to find the component of it along another vector \vec{b}b , actually it is the dot product or scalar product of \vec{a}a and the unit vector of \vec{b}b along \vec{b}b , i.e., \hat{b}b^ .
Component of \vec{a}a along \vec{b}b
= \vec{a}.\hat{b}a.b^ , where \hat{b}=\frac{\vec{b}}{|\vec{b}|}b^=∣b∣b
Solution:
1. The given vectors are
\vec{A}=2\hat{i}+3\hat{j}A=2i^+3j^
\vec{B}=\hat{i}+\hat{j}B=i^+j^
Now \hat{B}=\frac{\vec{B}}{|\vec{B}|}B^=∣B∣B
= \frac{\hat{i}+\hat{j}}{\sqrt{1^{2}+1^{2}}}12+12i^+j^
= \frac{1}{\sqrt{2}}(\hat{i}+\hat{j})21(i^+j^)
Then the component of \vec{A}A along \vec{B}B is
= \vec{A}.\hat{B}A.B^
= \frac{1}{\sqrt{2}}(2\hat{i}+3\hat{j}). (\hat{i}+\hat{j})21(2i^+3j^).(i^+j^)
= \frac{1}{\sqrt{2}}(2+3)21(2+3)
= \frac{5}{\sqrt{2}}25
2. The given vectors are
\vec{A}=2\hat{i}+3\hat{j}A=2i^+3j^
\vec{C}=\hat{i}-\hat{j}C=i^−j^
Now \hat{C}=\frac{\vec{C}}{|\vec{C}|}C^=∣C∣C
= \frac{\hat{i}-\hat{j}}{\sqrt{1^{2}+(-1)^{2}}}12+(−1)2i^−j^
= \frac{1}{\sqrt{2}}(\hat{i}-\hat{j})2