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Question:

The magnitude of the projection of the vector 2i^+3j^+k^ on the vector perpendicular to the plane containing the vectors i^+j^+k^ and i^+2j^+3k^ is

32

32

6

36

Solution:

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 = (cn) / ||n|| * (n/||n||)

where ||n|| is the magnitude of n.

cn = (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| = |(cn) / ||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 is the dot product of c and n
  • ||n|| is the magnitude of n

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.