Solveeit Logo
Login

Question

Mathematics Question on Vector Algebra

Find a vector in a direction of vector 5i^−j^+2k^5\hat{i}-\hat{j}+2\hat{k} which has magnitude 8units.

Answer

The correct answer is:=4030i⃗−8√30j⃗+16√30k⃗=\frac{40}{\sqrt{30}}\vec{i}\frac{-8}{√30}\vec{j}\frac{+16}{√30}\vec{k}
Let a⃗=5i^−j^+2k^\vec{a}=5\hat{i}-\hat{j}+2\hat{k}
∴∣a⃗∣=52+(−1)2+22=25+1+4=30∴|\vec{a}|=\sqrt{5^2+(-1)^2+2^2}=\sqrt{25+1+4}=\sqrt{30}
∴a^=a⃗∣a⃗∣=5i^−j^+2k^30∴\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{5\hat{i}-\hat{j}+2\hat{k}}{\sqrt{30}}
Hence,the vector in the direction of vector 5i^−j^+2k^5\hat{i}-\hat{j}+2\hat{k} which has magnitude 8units is given by,
8a^=8(5i^−j^+2k^)308\hat{a}=\frac{8(5\hat{i}-\hat{j}+2\hat{k})}{\sqrt{30}}
=4030i^−830j^+1630k^=\frac{40}{\sqrt{30}}\hat{i}-\frac{8}{\sqrt{30}}\hat{j}+\frac{16}{\sqrt{30}}\hat{k}
=8(5i⃗−j⃗+2k⃗30)=8(\frac{5\vec{i}-\vec{j}+2\vec{k}}{\sqrt{30}})
=4030i⃗−8√30j⃗+16√30k⃗=\frac{40}{\sqrt{30}}\vec{i}\frac{-8}{√30}\vec{j}\frac{+16}{√30}\vec{k}