Question

What is the component of (3i+4j) perpendicular to (i+j)?​

Answer 1

Answer:

Let us first find the angle between the two given vectors -

Cos(A)=(i+j)∙(3i+4j)√[12+12]√[32+42]

Cos(A)=75√2

Sin2(A)=1−Cos2(A)=1−4950

Sin(A)=15√2 …. (Considering the acute angle, so taken only +ve value)

Now let r=3i+4j . This vector has a magnitude of |r|= √[32+42]=5 .

If we resolve r into two components, one parallel to i+j and other perpendicular to i+j , then since the acute angle between these two given vectors is A , so the magnitude of parallel component is |r| Cos(A) and the perpendicular component is |r| Sin(A)

Since the questions demands the perpendicular component, we need to multiply this magnitude with a unit vector perpendicular to i+j , which is

− 1√2i+1√2j

So the required answer is -

|r|Sin(A)(−1√2i+1√2j)

=515√2(−1√2i+1√2j)

=−12i+12j

The question demand that the result must be in the same plane as 3i+4j , which it is given that the entire problem and solution is present only in 2D - X-Y plane.

Let us check. If the answer is indeed perpendicular to i+j , their dot product must be zero

(−12i+12j)∙(i+j)

−12+12=0