32
32
6
36
Let the given vectors be
a = i^+j^+k^
b = i^+2j^+3k^
c = 2i^+3j^+k^
A vector perpendicular to the plane containing a and b is given by their cross product:
n = a x b = (i^+j^+k^) x (i^+2j^+3k^) = i^(3-2) - j^(3-1) + k^(2-1) = i^-2j^+k^
The projection of vector c on n is given by:
projn c = (c • n) / ||n|| * (n/||n||)
where ||n|| is the magnitude of n.
c • n = (2i^+3j^+k^) • (i^-2j^+k^) = 2 - 6 + 1 = -3
||n|| = √(1²+(-2)²+1²) = √6
Therefore, the magnitude of the projection is:
|projn c| = |(c • n) / ||n|| | = |-3/√6| = 3/√6 = √6/2 ≈ 1.225
However, none of the given options match this result. There might be an error in the question or the provided options. Let's re-examine the calculation. The magnitude of the projection of vector c onto the normal vector n is given by:
|projn c| = |c . n| / ||n||
Where:
c . n = (2)(1) + (3)(-2) + (1)(1) = 2 - 6 + 1 = -3
||n|| = √(1² + (-2)² + 1²) = √6
|projn c| = |-3| / √6 = 3/√6 = √6/2
Let's rationalize the denominator:
(√6/2) * (√6/√6) = 6/ (2√6) = 3/√6 = √6/2 ≈ 1.2247
None of the options match this value. There must be a mistake in the question or the options provided.