NIMCET 2021 Mathematics PYQ — Let , and are three vectors. Then, a vector in the plane of and w… | Mathem Solvex | Mathem Solvex
Tip:A–D to answerE for explanationV for videoS to reveal answer
NIMCET 2021 — Mathematics PYQ
NIMCET | Mathematics | 2021
Let a=2i^+j^+2k^, b=i^−j^+2k^ and c=i^+j^−2k^ are three vectors. Then, a vector in the plane of a and c whose projection on b is of magnitude 61 is
Choose the correct answer:
A.
3i^−2j^
B.
3i^+2j^
(Correct Answer)
C.
2i^+3j^−k^
D.
3i^+2j^+k^
Correct Answer:
3i^+2j^
Explanation
1. Define the given vectors: The vectors are given as a=2i^+j^+2k^, b=i^−j^+2k^, and c=i^+j^−2k^. [1, 2] 1. Express a vector in the plane of a and c: A vector r in the plane of a and c can be expressed as a linear combination of a and c, i.e., r=λa+μc, where λ and μ are scalars. Substituting the given vectors: r=λ(2i^+j^+2k^)+μ(i^+j^−2k^)r=(2λ+μ)i^+(λ+μ)j^+(2λ−2μ)k^. [1] 1. Calculate the projection of r on b: The projection of r on b is given by ∣b∣r⋅b. First, calculate the magnitude of b: ∣b∣=(1)2+(−1)2+(2)2=1+1+4=6. Next, calculate the dot product r⋅b: r⋅b=((2λ+μ)i^+(λ+μ)j^+(2λ−2μ)k^)⋅(i^−j^+2k^)r⋅b=(2λ+μ)(1)+(λ+μ)(−1)+(2λ−2μ)(2)r⋅b=2λ+μ−λ−μ+4λ−4μr⋅b=5λ−4μ. The projection of r on b is 65λ−4μ. [1] 1. Set the magnitude of the projection equal to 61: It is given that the magnitude of the projection is 61. ∣65λ−4μ∣=61. This implies ∣5λ−4μ∣=1. So, 5λ−4μ=1 or 5λ−4μ=−1. 1. Test the given options: The options are vectors. We need to find which option satisfies the condition. Let's assume the vector is r=xi^+yj^+zk^. Comparing this with r=(2λ+μ)i^+(λ+μ)j^+(2λ−2μ)k^, we have: x=2λ+μy=λ+μz=2λ−2μ Let's check option (b) 3i^+2j^: Here, x=3, y=2, z=0. 2λ+μ=3 (Equation 1) λ+μ=2 (Equation 2) 2λ−2μ=0⟹λ=μ (Equation 3) Substitute λ=μ into Equation 2: λ+λ=2⟹2λ=2⟹λ=1. Since λ=μ, then μ=1. Now, check if these values satisfy Equation 1: 2(1)+1=3, which is true. Finally, check the projection condition: 5λ−4μ=5(1)−4(1)=5−4=1. This satisfies the condition ∣5λ−4μ∣=1. Final Answer The vector in the plane of a and c whose projection on b is of magnitude 61 is \boxed{\text{3i ̂+2j ̂}}.
Explanation
1. Define the given vectors: The vectors are given as a=2i^+j^+2k^, b=i^−j^+2k^, and c=i^+j^−2k^. [1, 2] 1. Express a vector in the plane of a and c: A vector r in the plane of a and c can be expressed as a linear combination of a and c, i.e., r=λa+μc, where λ and μ are scalars. Substituting the given vectors: r=λ(2i^+j^+2k^)+μ(i^+j^−2k^)r=(2λ+μ)i^+(λ+μ)j^+(2λ−2μ)k^. [1] 1. Calculate the projection of r on b: The projection of r on b is given by ∣b∣r⋅b. First, calculate the magnitude of b: ∣b∣=(1)2+(−1)2+(2)2=1+1+4=6. Next, calculate the dot product r⋅b: r⋅b=((2λ+μ)i^+(λ+μ)j^+(2λ−2μ)k^)⋅(i^−j^+2k^)r⋅b=(2λ+μ)(1)+(λ+μ)(−1)+(2λ−2μ)(2)r⋅b=2λ+μ−λ−μ+4λ−4μr⋅b=5λ−4μ. The projection of r on b is 65λ−4μ. [1] 1. Set the magnitude of the projection equal to 61: It is given that the magnitude of the projection is 61. ∣65λ−4μ∣=61. This implies ∣5λ−4μ∣=1. So, 5λ−4μ=1 or 5λ−4μ=−1. 1. Test the given options: The options are vectors. We need to find which option satisfies the condition. Let's assume the vector is r=xi^+yj^+zk^. Comparing this with r=(2λ+μ)i^+(λ+μ)j^+(2λ−2μ)k^, we have: x=2λ+μy=λ+μz=2λ−2μ Let's check option (b) 3i^+2j^: Here, x=3, y=2, z=0. 2λ+μ=3 (Equation 1) λ+μ=2 (Equation 2) 2λ−2μ=0⟹λ=μ (Equation 3) Substitute λ=μ into Equation 2: λ+λ=2⟹2λ=2⟹λ=1. Since λ=μ, then μ=1. Now, check if these values satisfy Equation 1: 2(1)+1=3, which is true. Finally, check the projection condition: 5λ−4μ=5(1)−4(1)=5−4=1. This satisfies the condition ∣5λ−4μ∣=1. Final Answer The vector in the plane of a and c whose projection on b is of magnitude 61 is \boxed{\text{3i ̂+2j ̂}}.
