Question

If vector a = 2i+3j+6k and b=3i+4j then find projection of a on b also find projection of b on a

Answer 1

Answer:

The projection of $\vec{a}$ on $\vec{b}$ is 3.

The projection of $\vec{b}$ on $\vec{a}$ is $\frac{15}{7}$.

Step-by-step explanation:

Given : If vector a = 2i+3j+6k and b=3i+4j.

To find : Projection of a on b also projection of b on a?

Solution :

The formula of projection of $\vec{a}$ on $\vec{b}$ is $=\frac{1}{|\vec{b}|}(\vec{a}\cdot\vec{b})$

a = 2i+3j+6k and b=3i+4j

$\vec{a}\cdot\vec{b}=2(3)+3(3)+6(0)$

$\vec{a}\cdot\vec{b}=6+9+0$

$\vec{a}\cdot\vec{b}=15$

Magnitude of b,

$|\vec{b}|=\sqrt{3^2+4^2}$

$|\vec{b}|=\sqrt{9+16}$

$|\vec{b}|=\sqrt{25}$

$|\vec{b}|=5$

Substitute the value in the formula,

Projection of $\vec{a}$ on $\vec{b}$ is $=\frac{1}{|\vec{b}|}(\vec{a}\cdot\vec{b})$

$=\frac{1}{5}(15)$

$=3$

The projection of $\vec{a}$ on $\vec{b}$ is 3.

The formula of projection of $\vec{b}$ on $\vec{a}$ is $=\frac{1}{|\vec{a}|}(\vec{a}\cdot\vec{b})$

a = 2i+3j+6k and b=3i+4j

$\vec{a}\cdot\vec{b}=15$

Magnitude of a,

$|\vec{a}|=\sqrt{2^2+3^2+6^2}$

$|\vec{a}|=\sqrt{4+9+36}$

$|\vec{a}|=\sqrt{49}$

$|\vec{a}|=7$

Substitute the value in the formula,

Projection of $\vec{b}$ on $\vec{a}$ is $=\frac{1}{|\vec{b}|}(\vec{a}\cdot\vec{b})$

$=\frac{1}{7}(15)$

$=\frac{15}{7}$

The projection of $\vec{b}$ on $\vec{a}$ is $\frac{15}{7}$.

Answer 2

Answer:

15/27

Step-by-step explanation:

15/27. .jdjdjdjdjchhdudjdndnsnsmsk