Let overline{a}, overline{b}, overline{c} be | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We know that the scalar triple product $[\overline{a} \overline{b} \overline{c}] = (\overline{a} 	imes \overline{b}) \cdot \overline{c}$.For the

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) We know that the scalar triple product $[\overline{a} \overline{b} \overline{c}] = (\overline{a} 	imes \overline{b}) \cdot \overline{c}$.For the first term: $\overline{p} \cdot (\overline{a} + \overline{b}) = \overline{p} \cdot \overline{a} + \overline{p} \cdot \overline{b}$.Substituting $\overline{p} = \frac{\overline{b} 	imes \overline{c}}{[\overline{a} \overline{b} \overline{c}]}$,we get:$\overline{p} \cdot \overline{a} = \frac{(\overline{b} 	imes \overline{c}) \cdot \overline{a}}{[\overline{a} \overline{b} \overline{c}]} = \frac{[\overline{b} \overline{c} \overline{a}]}{[\overline{a} \overline{b} \overline{c}]} = \frac{[\overline{a} \overline{b} \overline{c}]}{[\overline{a} \overline{b} \overline{c}]} = 1$.$\overline{p} \cdot \overline{b} = \frac{(\overline{b} 	imes \overline{c}) \cdot \overline{b}}{[\overline{a} \overline{b} \overline{c}]} = 0$ (since $\overline{b} 	imes \overline{c}$ is perpendicular to $\overline{b}$).Thus,$\overline{p} \cdot (\overline{a} + \overline{b}) = 1 + 0 = 1$.Similarly,$\overline{q} \cdot (\overline{b} + \overline{c}) = 1$ and $\overline{r} \cdot (\overline{c} + \overline{a}) = 1$.Adding these,we get $1 + 1 + 1 = 3$.

Similar Questions

Let the volume of tetrahedron $ABCD$ be $81$ cubic units and $G_1, G_2, G_3$ be the centroids of the triangular faces $ABC, ABD,$ and $ACD$ respectively. Then the volume of tetrahedron $AG_1G_2G_3$ is (in cubic units):

If $\vec{u} = \hat{j} + 4\hat{k}$,$\vec{v} = \hat{i} + 3\hat{k}$ and $\vec{w} = \cos \theta \hat{i} + \sin \theta \hat{j}$ are vectors in $3$-dimensional space,then the maximum possible value of $|(\vec{u} \times \vec{v}) \cdot \vec{w}|$ is

If the given vectors $(-bc, b^2 + bc, c^2 + bc)$,$(a^2 + ac, -ac, c^2 + ac)$ and $(a^2 + ab, b^2 + ab, -ab)$ are coplanar,where none of $a, b$ and $c$ is zero,then:

The value of $[\vec{a}-\vec{b} \quad \vec{b}-\vec{c} \quad \vec{c}-\vec{a}]$ is equal to

If $A$,$B$,and $C$ are three non-coplanar vectors,then $(A + B + C) \cdot ((A + B) \times (A + C)) = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module