Question

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

Answer 1

Answer:

Explanation:

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√2jSo 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