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Question

Consider the parallelopiped with sides a=3^i+2^j+^k,b=^i+^j+2^k and c=^i+3^j+3^k, then angle between a and the plane containing the face determined by b and c is

A
sin113
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B
cos1914
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C
sin1914
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D
sin123
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Solution

The correct option is C sin1914
¯¯bׯ¯c:∣ ∣ ∣ijk112133∣ ∣ ∣
i(36)j(32)+k(31)
3ij+2k
Ifθistheanglebetween¯¯¯aandtheplane
containing¯¯band¯¯c,then:
cos(90θ)=∣ ∣¯¯¯a.(¯¯bׯ¯c)¯¯¯a.¯¯bׯ¯c∣ ∣
(90θ)willbetheanglewiththenormal
totheplane.
¯¯¯a.(¯¯bׯ¯c)=(3i+2j+k).(3ij+2k)
=92+2
=9
¯¯¯a=32+22+1=14
¯¯bׯ¯c=3ij+2k
=32+12+4
=9+1+4=14
¯¯¯a.¯¯bׯ¯c=(14)2=14
∣ ∣¯¯¯a.(¯¯bׯ¯c)¯¯¯a.¯¯bׯ¯c∣ ∣=914
cos(90θ)=914
sinθ=914
θ=sin1(914)

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