Question

A vector with magnitude of 3 minutes which is perpendicular to each of the vectors a is equal to 4i + 2j - 6k b is equal to 8i + 6j + 9 k is given by

Answer 1

± (81𝑖 − 126𝑗 + 12𝑘)/√2509

Answer 2

We need a vector v of magnitude 3 that is perpendicular to both

a = 4𝑖 + 2𝑗 − 6𝑘

b = 8𝑖 + 6𝑗 + 9𝑘.

Step 1. Compute the cross product a × b:

a × b = | 𝑖 𝑗 𝑘 | | 4 2 -6 | | 8 6 9 |

Expanding this determinant gives:

𝑖 (2·9 – (–6·6)) − 𝑗 (4·9 – (–6·8)) + 𝑘 (4·6 – 2·8) = 𝑖 (18 + 36) − 𝑗 (36 + 48) + 𝑘 (24 – 16) = 54𝑖 − 84𝑗 + 8𝑘.

Step 2. The magnitude of a × b is:

|a × b| = √(54² + (–84)² + 8²) = √(2916 + 7056 + 64) = √10036 = 2√2509.

Step 3. The unit vector perpendicular to both is:

u = (1/(2√2509)) (54𝑖 − 84𝑗 + 8𝑘).

Step 4. Scaling to magnitude 3, we have:

v = 3u = (3/(2√2509)) (54𝑖 − 84𝑗 + 8𝑘).

Notice that 54, 84, and 8 are all even; factor 2 out: (54, –84, 8) = 2(27, –42, 4).

Thus,

v = (3/(2√2509)) · 2(27𝑖 − 42𝑗 + 4𝑘) = (3/(√2509)) (27𝑖 − 42𝑗 + 4𝑘).

So the required vector is:

v = ± (81𝑖 − 126𝑗 + 12𝑘)/√2509 (since 3×27=81, 3×42=126, 3×4=12).