If a is a nonzero vector such that its projections on the vectors 2i^−j^+2k^, i^+2j^−2k^, and k^ are equal, then a unit vector along a is:
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Step-by-Step Solution
Step 1: Define the given vectors and projection formula
We are given three vectors, let's call them b1, b2, and b3. The problem states that the projection of an unknown vector a onto each of these three vectors is equal. We recall the formula for the projection of vector a onto vector b, which is given by the dot product of a and b divided by the magnitude of b.
Step 2: Calculate magnitudes of the given vectors
To use the projection formula, we need the magnitudes of the vectors b1, b2, and b3. We calculate these magnitudes using the formula ∣v∣=vx2+vy2+vz2. We find that ∣b1∣ is 3, ∣b2∣ is 3, and ∣b3∣ is 1.
Step 3: Set up equations based on equal projections
The problem states that the projections are equal. We set up the equality using the projection formula and the calculated magnitudes. Let a=xi^+yj^+zk^.
Step 4: Solve for the components of vector a
We equate the first two projections and the second and third projections to form a system of linear equations. Solving these equations for x, y, z in terms of a common ratio, we find that x, y, and z are proportional to 7, 9, and 5 respectively. This means a can be written as k(7i^+9j^+5k^) for some scalar k.
Step 5: Find the unit vector along a
Since a is a nonzero vector, k=0. The unit vector along a is found by dividing a by its magnitude. This gives us two possible unit vectors, differing only by a sign, as the problem does not specify a direction for a.