What is the component of 3i+4j perpendicular to i+j ans in the same plane as that of 3i+4j?
Answers
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√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
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