Question

A parallelogram is constructed on the vectors $\overline{\alpha }$ and $\overline{\beta }$. A vector which coincides with the altitude of the parallelogram and perpendicular to the side $\overline{\alpha }$ expressed in terms of the vectors $\overline{\alpha }$ and $\overline{\beta }$ is

• A. $\overline{\beta }+\frac{\overline{\beta }-\overline{\alpha }}{{\left(\overline{\alpha }\right)}^2}\overline{\alpha }$
• B. $\frac{(\overline{\alpha }\times \overline{\beta })\times \overline{\alpha }}{{\left(\overline{\alpha }\right)}^2}$
• C. $\frac{\overline{\beta }\cdot \overline{\alpha }}{{\left(\alpha \right)}^2}\overline{\alpha }+\overline{\beta }$
• D. $\left|\overline{\beta }\right|\frac{\overline{\alpha }\times (\overline{\alpha }\times \overline{\beta })}{{\left(\alpha \right)}^2}$

Answer 1

The correct option is B $\frac{(\overline{\alpha }\times \overline{\beta })\times \overline{\alpha }}{{\left(\overline{\alpha }\right)}^2}$

$(\alpha \times \beta )$ will give us the perpendicular vector or the normal to the plane where the parallelogram is lying.

Now

$(\alpha \times \beta )\times \alpha$ will give the vector parallel to the altitude of the parallelogram which is perpendicular to side $\overline{\alpha }$.

Thus the unit vector of the altitude of the given parallelogram perpendicular to $\overline{\alpha }$ will be

$=\frac{(\alpha \times \beta )\times \alpha }{|\alpha {|}^2}$