JEE 2022 Mathematics PYQ — Let , and where , be three vectors. If the projection of on is an… | Mathem Solvex | Mathem Solvex
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JEE 2022 — Mathematics PYQ
JEE | Mathematics | 2022
Let a=αi^+3j^−k^, b=3i^−βj^+4k^ and c=i^+2j^−2k^ where α,β∈R, be three vectors. If the projection of a on c is 310 and b×c=−6i^+10j^+7k^, then the value of α+β is equal to:
Choose the correct answer:
A.
3
(Correct Answer)
B.
4
C.
5
D.
6
Correct Answer:
3
Explanation
Solution
Step 1: Find the value of α using the projection formula.
The projection of a on c is given by:
Projection=∣c∣a⋅c
Given ∣c∣=12+22+(−2)2=1+4+4=3.
According to the question:
3(αi^+3j^−k^)⋅(i^+2j^−2k^)=310
α(1)+3(2)+(−1)(−2)=10
α+6+2=10
α+8=10⟹α=2
Step 2: Find the value of β using the cross product.
Given b×c=−6i^+10j^+7k^. We calculate the cross product using the determinant method:
b×c=i^31amp;j^amp;−βamp;2amp;k^amp;4amp;−2
b×c=i^(2β−8)−j^(−6−4)+k^(6+β)
b×c=(2β−8)i^+10j^+(6+β)k^
Comparing the coefficients of k^ (or i^) with the given vector −6i^+10j^+7k^:
6+β=7⟹β=1
(Checking with i^: 2(1)−8=−6, which matches).
Step 3: Calculate α+β.
α+β=2+1=3
Final Answer:
The value is 3. The correct option is (A).
Explanation
Solution
Step 1: Find the value of α using the projection formula.
The projection of a on c is given by:
Projection=∣c∣a⋅c
Given ∣c∣=12+22+(−2)2=1+4+4=3.
According to the question:
3(αi^+3j^−k^)⋅(i^+2j^−2k^)=310
α(1)+3(2)+(−1)(−2)=10
α+6+2=10
α+8=10⟹α=2
Step 2: Find the value of β using the cross product.
Given b×c=−6i^+10j^+7k^. We calculate the cross product using the determinant method:
b×c=i^31amp;j^amp;−βamp;2amp;k^amp;4amp;−2
b×c=i^(2β−8)−j^(−6−4)+k^(6+β)
b×c=(2β−8)i^+10j^+(6+β)k^
Comparing the coefficients of k^ (or i^) with the given vector −6i^+10j^+7k^:
