Find α so vectors to (3,4,0) and (α,4,2) are perpendicular

[subzero0137]

Vectors a and b correspond to the vectors from the origin to the points A with co-ordinates (3,4,0) and B with co-ordinates (α,4, 2) respectively. Find a value of α that makes the scalar product a$\cdot$b equal to zero, and explain the physical significance.


My attempt:
The scalar product a$\cdot$b is given by |a||b|cosθ=$5 \sqrt{α^{2}+20}$cosθ=0, therefore $α=\sqrt{20}$i. But I don't know the physical significance of this. Please help!

 

[jbunniii]

No, $\alpha$ is a real number, so you won't be able to achieve $\sqrt{\alpha^2 + 20} = 0$. The solution you are seeking will give you $\cos \theta = 0$. But since you haven't related $\theta$ to $\alpha$, that doesn't help much. Instead of using $a \cdot b = |a||b|\cos \theta$, do you know another formula for $a \cdot b$?

 

[HallsofIvy]

Vectors a and b correspond to the vectors from the origin to the points A with co-ordinates (3,4,0) and B with co-ordinates (α,4, 2) respectively. Find a value of α that makes the scalar product a$\cdot$b equal to zero, and explain the physical significance.

My attempt:
The scalar product a$\cdot$b is given by |a||b|cosθ=$5 \sqrt{α^{2}+20}$cosθ=0

That is one way calculate the dot product but, rather than calculate $\theta$, it is simpler to use $(a, b, c)\cdot (u, v, w)= au+ bv+ cw$. Here that would be 3a+ 16+ 0= 3a+ 16= 0.

, therefore $α=\sqrt{20}$i.

No, a must be a real number. The fact that the dot product is 0 does NOT mean one of the vectors must have length 0. It is also possible that $cos(\theta)= 0$.

But I don't know the physical significance of this. Please help!

Two non-zero vectors have 0 dot product if and only if they are perpendicular.