Question

Find the projection of A on B. Given A = 10j + 3k and B = 4j + 5k

Answer 1

Answer:

Given A = 10j + 3k and B = 4j + 5k. Explanation: Projection of A on B = (A . B)/|B|. Thus the answer is 40/6.4= 6.25.

Answer 2

Answer:

$\frac{55}{\sqrt{41} }$ is the projection of A on B.

Step-by-step explanation:

Explanation:

Given, A = $10\vec{j} +3\vec{k}$ and

B = $4\vec{j} +5\vec{k}$

As we know to find the projection of A on B by using the method

$\frac{\vec{a}.\vec{b}}{| b |}$ .

Step1:

We have , A = $10\vec{j} +3\vec{k}$ and

B = $4\vec{j} +5\vec{k}$

Therefore , $\vec{A}.\vec{B} = (10\vec{j} +3\vec{k}). (4\vec{j} +5\vec{k})$

                  ⇒$\vec{A}.\vec{B} = (40+15)$

                   ⇒  $\vec{A}.\vec{B} = 55$

Now we find | B |  

We have B = $4\vec{j} +5\vec{k}$

 Therefore , | A |= $\sqrt{4^{2} +5^{2} }$ = $\sqrt{41}$

Step2:

So , the projection of A on B  = $\frac{\vec{A}.\vec{B}}{|B|}$

Now put the value of |B | = $\sqrt{41}$ and $\vec{A}.\vec{B} = 55$ in thr above formula .

⇒ $\frac{\vec{A}.\vec{B}}{|B|} = \frac{55}{\sqrt{41} }$ .

Final answer :

Hence , the projection of A on B is $\frac{55}{\sqrt{41} }$ .