Let $u, v$ and $w$ are non-coplanar vectors, then the value of $\frac{(u+2 v-w) \cdot[(u-v)…

Let $u, v$ and $w$ are non-coplanar vectors, then the value of $\frac{(u+2 v-w) \cdot[(u-v) \times(u-v-w)]}{[u v w]}$
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Solution

Given u, v and w are non coplanar vectors. To Find The value of $\frac{(u+2 v-w) \cdot[u-v) \times(u-v-w)]}{[u v w]}$ $\begin{aligned} & (u+2 v-w) \cdot[u \times u-u \times v-u \times w \\ & \frac{-v \times u+v \times v+v \times w]}{[u v w]} \\ & \end{aligned}$ $\begin{aligned} & =\frac{(u+2 v-w) \cdot[0-u \times v-u \times w+u \times v+0+v \times w]}{[u v w]} \\ & u \cdot(-u \times w)+u \cdot(v \times w)-2 v(u \times w)+2 v \cdot(v \times w) \\ & =\frac{+w \cdot(u \times w)-w \cdot(v \times w)}{[u v w]} \\ & \end{aligned}$ $\begin{aligned} & -\left[\begin{array}{lll}u & u & w\end{array}\right]+\left[\begin{array}{lll}u & v & w\end{array}\right]-2\left[\begin{array}{lll}v & u & w\end{array}\right]+2\left[\begin{array}{lll}v & v & w\end{array}\right] \\ & =\frac{+[w u w]-[w \vee w]}{[u v w]} \\ & =\frac{[u \vee w]+2[u \vee w]}{[u \vee w]} \quad\{\because[u u w]=[v \vee w] \\ & \end{aligned}$ $\begin{aligned} & \left.=\left[\begin{array}{lll}w & u & w\end{array}\right]=\left[\begin{array}{lll}w & v & w\end{array}\right]=0 \text { and }\left[\begin{array}{lll}v & u & w\end{array}\right]=-\left[\begin{array}{lll}u & v & w\end{array}\right]\right\} \\ & =\frac{3[u \vee w]}{[u \vee w]}=3 \\ & \end{aligned}$

Asked in: AP EAMCET 2021 (25 Aug Shift 2)

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