NIMCET 2023 Mathematics PYQ — Let and . If is a vector such that and the angle between and is ,… | Mathem Solvex | Mathem Solvex
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NIMCET 2023 — Mathematics PYQ
NIMCET | Mathematics | 2023
Let A=2i^+j^−2k^ and B=i^+j^. If C is a vector such that ∣C−A∣=3 and the angle between A×B and C is 30∘, then [(A×B)×C]=3. Then, the value of A⋅C is equal to
Choose the correct answer:
A.
25/8
B.
2
(Correct Answer)
C.
5
D.
1/8
Correct Answer:
2
Explanation
1. Calculate A×B and its Magnitude
Given:
A=2i^+j^−2k^
B=i^+j^+0k^
Compute the cross product using determinants:
A×B=i^21amp;j^amp;1amp;1amp;k^amp;−2amp;0
A×B=i^(0−(−2))−j^(0−(−2))+k^(2−1)
A×B=2i^−2j^+k^
Now, find its magnitude:
∣A×B∣=22+(−2)2+12=4+4+1=9=3
2. Find the Magnitude of Vector C
We are given the cross product magnitude condition:
∣(A×B)×C∣=3
Using the cross product magnitude formula ∣X×Y∣=∣X∣∣Y∣sinθ, where θ=30∘ is the angle between (A×B) and C:
∣A×B∣⋅∣C∣⋅sin30∘=3
Substitute the known values (∣A×B∣=3 and sin30∘=21):
3⋅∣C∣⋅21=3
21∣C∣=1⟹∣C∣=2
3. Use the Magnitude Condition to Find A⋅C
We are given:
∣C−A∣=3
Square both sides of the equation:
∣C−A∣2=32
∣C∣2+∣A∣2−2(A⋅C)=9
We need the magnitude of A:
∣A∣=22+12+(−2)2=4+1+4=3⟹∣A∣2=9
Substitute ∣C∣=2 and ∣A∣2=9 into the squared equation:
22+9−2(A⋅C)=9
4+9−2(A⋅C)=9
Subtract 9 from both sides:
4−2(A⋅C)=0
2(A⋅C)=4
A⋅C=2
Explanation
1. Calculate A×B and its Magnitude
Given:
A=2i^+j^−2k^
B=i^+j^+0k^
Compute the cross product using determinants:
A×B=i^21amp;j^amp;1amp;1amp;k^amp;−2amp;0
A×B=i^(0−(−2))−j^(0−(−2))+k^(2−1)
A×B=2i^−2j^+k^
Now, find its magnitude:
∣A×B∣=22+(−2)2+12=4+4+1=9=3
2. Find the Magnitude of Vector C
We are given the cross product magnitude condition:
∣(A×B)×C∣=3
Using the cross product magnitude formula ∣X×Y∣=∣X∣∣Y∣sinθ, where θ=30∘ is the angle between (A×B) and C:
∣A×B∣⋅∣C∣⋅sin30∘=3
Substitute the known values (∣A×B∣=3 and sin30∘=21):
3⋅∣C∣⋅21=3
21∣C∣=1⟹∣C∣=2
3. Use the Magnitude Condition to Find A⋅C
We are given:
∣C−A∣=3
Square both sides of the equation:
∣C−A∣2=32
∣C∣2+∣A∣2−2(A⋅C)=9
We need the magnitude of A:
∣A∣=22+12+(−2)2=4+1+4=3⟹∣A∣2=9
Substitute ∣C∣=2 and ∣A∣2=9 into the squared equation:
