If $\vec{a} = \hat{i} - \hat{k}$, $\vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}$ and $\vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k}$, then the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ depends on
- Aonly $x$
- Bneither $x$ nor $y$
- Ceither $x$ or $y$
- Donly $y$
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Similar Questions
If $\bar{a}$ is perpendicular to $\bar{b}$ and $\bar{c}$,$|\vec{a}|=2$,$|\bar{b}|=3$,$|\bar{c}|=4$ and the angle between $\bar{b}$ and $\bar{c}$ is $\frac{\pi}{3}$,then $\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]=$ (in $\sqrt{3}$)
If $\bar{a}, \bar{b},$ and $\bar{c}$ are mutually perpendicular vectors such that $|\bar{a}| = 1, |\bar{b}| = 3,$ and $|\bar{c}| = 5,$ then find the value of $[\bar{a} - 2\bar{b}, \bar{b} - 3\bar{c}, \bar{c} - 4\bar{a}]$.
Difficult
View SolutionIf $\vec{a}, \vec{b}, \vec{c}$ are three non-coplanar vectors and $\vec{r}$ is any vector,then $[\vec{b} \, \vec{c} \, \vec{r}] \vec{a} + [\vec{c} \, \vec{a} \, \vec{r}] \vec{b} + [\vec{a} \, \vec{b} \, \vec{r}] \vec{c} = \dots$
Difficult
View SolutionLet $a=p(\hat{i}+\hat{j}+\hat{k})$, $b=\hat{i}+\hat{j}-2\hat{k}$, and $c=2\hat{i}-\hat{j}+2\hat{k}$ be three vectors. If the value of $[abc]$ is not more than $15$ and not less than $-5$, then $p$ lies in the interval:
$i \cdot (j \times k) + j \cdot (k \times i) + k \cdot (i \times j) = $
Easy
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