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Question : 69 of 120
Marks:
+1,
-0
Solution:
Given,
The position vectors of points
A,B, and
C are
,, and
respectively, and
=cos2θ+sin‌2θ.
The expression to evaluate is:
(×)+(×)+(×).
First, substitute
into the equation:
(×)+(×(cos2θ+sin‌2θ))+(cos2θ+sin‌2θ)×Using the distributive property of the cross product:
(×)+[(×cos2θ)+(×sin‌2θ)]+[(cos2θ×)+(sin‌2θ×)]Since
×=0 and
×=0, we are left with:
(×)+cos2θ(×)+sin‌2θ(−×)Substitute
×=−(×) into the expression:
(×)+cos2θ(−×)+sin‌2θ(−×)Factor out
× :
×[1−cos2θ−sin‌2θ]Since
cos2θ+sin‌2θ=1, the expression becomes:
×[1−1]=0∴ The final result is
.
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