A parallelogram is constructed on the vectors ¯α and ¯β. A vector which coincides with the altitude of the parallelogram and perpendicular to the side ¯α expressed in terms of the vectors ¯α and ¯β is
A
¯β+¯β−¯α(¯α)2¯α
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(¯αׯβ)ׯα(¯α)2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
¯β⋅¯α(α)2¯α+¯β
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
∣∣¯β∣∣¯α×(¯αׯβ)(α)2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B(¯αׯβ)ׯα(¯α)2 (α×β) will give us the perpendicular vector or the normal to the plane where the parallelogram is lying.
Now
(α×β)×α will give the vector parallel to the altitude of the parallelogram which is perpendicular to side ¯α.
Thus the unit vector of the altitude of the given parallelogram perpendicular to ¯α will be