(a+b) cdot (b+c) times (a+b+c) is equal to | vedclass

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

Question

(B) We need to evaluate the scalar triple product: $(a+b) \cdot ((b+c) 	imes (a+b+c))$.First, expand the cross product term: $(b+c) 	imes (a+b+c) =

Vedclass · Accepted Answer

$[a b c]$

Vedclass · Answer

(B) We need to evaluate the scalar triple product: $(a+b) \cdot ((b+c) 	imes (a+b+c))$.First, expand the cross product term: $(b+c) 	imes (a+b+c) = (b 	imes a) + (b 	imes b) + (b 	imes c) + (c 	imes a) + (c 	imes b) + (c 	imes c)$.Since the cross product of any vector with itself is zero ($b 	imes b = 0$ and $c 	imes c = 0$), and using the property $c 	imes b = -(b 	imes c)$, we get:$(b+c) 	imes (a+b+c) = (b 	imes a) + (b 	imes c) + (c 	imes a) - (b 	imes c) = (b 	imes a) + (c 	imes a)$.Now, take the dot product with $(a+b)$:$(a+b) \cdot ((b 	imes a) + (c 	imes a)) = a \cdot (b 	imes a) + a \cdot (c 	imes a) + b \cdot (b 	imes a) + b \cdot (c 	imes a)$.Using the properties of scalar triple products, $a \cdot (b 	imes a) = 0$, $a \cdot (c 	imes a) = 0$, and $b \cdot (b 	imes a) = 0$.This leaves us with $b \cdot (c 	imes a) = [b c a]$.Since $[b c a] = [a b c]$, the final result is $[a b c]$.

$(a+b) \cdot (b+c) \times (a+b+c)$ is equal to

Similar Questions

What are the metal ions present in carnallite?

$A$ particle is moving on a circular path of radius $r$ with uniform speed $v$. The change in velocity when the particle moves from $P$ to $Q$ is $(\angle POQ = 40^\circ)$.

In a stratified cambium,the fusiform initials are

If $A = \begin{bmatrix} -2 & 2 \\ -3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$,then $(B^{-1} A^{-1})^{-1}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module