Question

find the vector projection of 5i-4j+k along the vector 3i-2j+4k​

Answer 1

$\large{\bf{\gray{\underline{\underline{\orange{Given:}}}}}}$

$:\implies\sf\:\vec{A}=5\hat{i}-4\hat{j}+\hat{k}$

$:\implies\sf\:\vec{B}=3\hat{i}-2\hat{j}+4\hat{k}$

$\large{\bf{\gray{\underline{\underline{\green{To\:Find:}}}}}}$

We have to find vector projection of A along the vector B.

$\large{\bf{\gray{\underline{\underline{\pink{Solution:}}}}}}$

➳ Such questions can easily be solved by using the concept of dot product.

➳ Dot product of two vectros inclined at an angle Φ is given by

$\dag\:\boxed{\bf{\red{\vec{A}\:{\tiny{\bullet}}\:\vec{B}=AB\cos\phi}}}$

❂ Vector projection of A along the vector B is given by

$\circ\tt\:A\cos\phi=\dfrac{\vec{A}\:{\tiny{\bullet}}\:\vec{B}}{B}$

$\circ\tt\:A\cos\phi=\dfrac{(5\hat{i}-4\hat{j}+\hat{k})(3\hat{i}-2\hat{j}+4\hat{k})}{\sqrt{3^2+(-2)^2+4^2}}$

$\circ\tt\:A\cos\phi=\dfrac{15+8+4}{\sqrt{29}}$

$\circ\:\boxed{\bf{\gray{A\cos\phi=\dfrac{27}{\sqrt{29}}}}}$

Answer 2

Answer:

Explanation:

We have to find the vector projection not just the magnitude