Question

a1i +a2j is a unit vector perpendicular to 4i - 3j find a1 and a2

Answer 1

Answer:

Explanation:

If these 2 vectors are perpendicular, their Dot product must be Zero.

Which means 4(a1)-3(a2)=0

or we can write (a1) = 3(a2)/4.—————-(1)

Also, since (a1)i+(a2)j is a unit vector, it means -

sqrt[(a1*a1)+(a2*a2)] = 1, {sqrt = square root}

squaring both sides, we get

(a1)*(a1)+(a2)*(a2) = 1———————-(2)

Putting value of (a1) in eqn(2), we get -

[3(a2)/4]*[3(a2)/4] + (a2)*(a2) = 1.

Solving further, 25(a2)*(a2) = 16

Take square root both sides,

+-5(a2) = +-4

It gives 2 values of (a2) which are +4/5 & -4/5.

Correspondingly (a1) = 3(a2)/4 becomes +3/5 & -3/5.

Hence we obtain 2 solutions for the question,

(a1) = +3/5, (a2) = +4/5 and

(a1) = -3/5, (a2) = - 4/5.