Let a=i^+2j^+k^ and b=2i^+j^−k^. Let c be a unit vector in the plane of the vectors a and b and be perpendicular to a. Then such a vector c is:
Get the complete, step-by-step math solution for: "Let {a} = {i} + 2 {j} + {k} and {b} = 2 {i} + {j} - {k}. Let {c} be a unit vector in the plane of the vectors {a} and {b} and be perpendicular to {a}....". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Express vector c in the plane of a and b
Since vector c lies in the plane of vectors a and b, it can be expressed as a linear combination of a and b. Here, x and y are scalar coefficients.
Step 2: Use the perpendicularity condition
We are given that c is perpendicular to a. This means their dot product is zero. Substituting the expression for c, we get an equation involving x and y. First, let's calculate the dot products a⋅a and a⋅b.
Step 3: Calculate dot products and solve for x/y
We calculate a⋅a=(1)(1)+(2)(2)+(1)(1)=1+4+1=6. And a⋅b=(1)(2)+(2)(1)+(1)(−1)=2+2−1=3. Substituting these into the dot product equation, we get 6x+3y=0, which simplifies to y=−2x. (Correction: a⋅b=3, so 6x+3y=0⟹y=−2x). Let's recheck the calculation. a⋅b=(1)(2)+(2)(1)+(1)(−1)=2+2−1=3. So 6x+3y=0⟹y=−2x.
Step 4: Substitute y and find vector c
Now we substitute y=−2x back into the expression for c. This gives us c=xa−2xb=x(a−2b). Let's calculate the vector a−2b.
Step 5: Calculate a - 2b and its magnitude
We compute a−2b=(1−4)i^+(2−2)j^+(1−(−2))k^=−3i^+0j^+3k^=−3i^+3k^. The magnitude of this vector is (−3)2+32=9+9=18=32.
Step 6: Normalize vector c
Since c is a unit vector, its magnitude is 1. We have ∣c∣=∣x∣∣a−2b∣. Setting this to 1, we find the possible values for x.
Step 7: Final vector c
Substitute the value of x back into the expression for c. This gives us the two possible unit vectors that satisfy the given conditions. We can simplify the expression by factoring out 3 from the vector and canceling it with the 3 in the denominator.