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Question
If the two adjacent sides of two rectangles are represented by the vectors →p=5→a−3→b;→q=−→a−2→b and →r=−4→a−→b;→s=−→a+→b respectively, then the angle between the vectors →x=13(→p+→r+→s) and →y=15(→r+→s) is
If the two adjacent sides of two rectangles are represented by the vectors →p=5→a−3→b;→q=−→a−2→b and →r=−4→a−→b;→s=−→a+→b respectively, then the angle between the vectors →x=13(→p+→r+→s) and →y=15(→r+→s) is
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Solution
The correct option is B cos−1(195√43)
Adjacent sides of a rectangle are perpendicular.
Thus, (5→a−3→b)⋅(−→a−2→b)=0⇒−5a2−7→a⋅→b+6b2=0
Also, (−4→a−→b)⋅(−→a+→b)=0⇒4a2−3→a⋅→b−b2=0
Let ab=j and θ be the angle between them.
The above equations thus become −5j2−7jcosθ+6=0 & 4j2−3jcosθ−1=0
→x=→p+→r+→s3=−3→b3=−→b
→y=→r+→s5=−→a
Thus, angle between →x,→y is θ
We thus solve the two quadratics written above.
−15j2+18=28j2−7⇒j2=2543
⇒cosθ=4j2−13j=573×5×√43=195√43
Adjacent sides of a rectangle are perpendicular.
Thus, (5→a−3→b)⋅(−→a−2→b)=0⇒−5a2−7→a⋅→b+6b2=0
Also, (−4→a−→b)⋅(−→a+→b)=0⇒4a2−3→a⋅→b−b2=0
Let ab=j and θ be the angle between them.
The above equations thus become −5j2−7jcosθ+6=0 & 4j2−3jcosθ−1=0
→x=→p+→r+→s3=−3→b3=−→b
→y=→r+→s5=−→a
Thus, angle between →x,→y is θ
The above equations thus become −5j2−7jcosθ+6=0 & 4j2−3jcosθ−1=0
→x=→p+→r+→s3=−3→b3=−→b
→y=→r+→s5=−→a
Thus, angle between →x,→y is θ
We thus solve the two quadratics written above.
−15j2+18=28j2−7⇒j2=2543
⇒cosθ=4j2−13j=573×5×√43=195√43
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