Search any question & find its solution
Question:
Answered & Verified by Expert
If $\overrightarrow{\mathbf{a}}=\hat{\mathrm{i}}-\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=x \hat{\mathbf{i}}+\hat{\mathbf{j}}+(1-x) \hat{\mathbf{k}}$
$\overrightarrow{\mathbf{c}}=y \hat{\mathbf{i}}+\mathrm{x} \hat{\mathbf{j}}+(1+x-y) \hat{\mathbf{k}}$
then $\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})$ depends on
Options:
$\overrightarrow{\mathbf{c}}=y \hat{\mathbf{i}}+\mathrm{x} \hat{\mathbf{j}}+(1+x-y) \hat{\mathbf{k}}$
then $\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})$ depends on
Solution:
2513 Upvotes
Verified Answer
The correct answer is:
Neither x nor $y$
Let $\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}$
and $\vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$
$\begin{array}{l}\text { Now, }(\vec{b} \times \vec{c})=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ x & 1 & (1-x) \\ y & x & (1+x-y)\end{array}\right| \\ \begin{aligned}=& i\left(1+x-y-x+x^{2}\right)-\hat{j}\left(x+x^{2}-x y-y\right) \\+& k\left(x^{2}-y\right) \end{aligned} \\ =\hat{i}\left(1-y+x^{2}\right)-\hat{j}\left(x+x^{2}-x y-y\right)+\hat{k}\left(x^{2}-y\right) \\ \text { Now, } \vec{a} .(\vec{b} \times \vec{c})=1\left(1-y+x^{2}\right)+0\left(x+x^{2}-x y-y\right) \\ -1\left(x^{2}-y\right) \\ =1-y+x^{2}-x^{2}+y & \end{array}$
$=1$ which shows that $\vec{a} .(\vec{b} \times \vec{c})$ does not depend
on $x$ and $y$.
and $\vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$
$\begin{array}{l}\text { Now, }(\vec{b} \times \vec{c})=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ x & 1 & (1-x) \\ y & x & (1+x-y)\end{array}\right| \\ \begin{aligned}=& i\left(1+x-y-x+x^{2}\right)-\hat{j}\left(x+x^{2}-x y-y\right) \\+& k\left(x^{2}-y\right) \end{aligned} \\ =\hat{i}\left(1-y+x^{2}\right)-\hat{j}\left(x+x^{2}-x y-y\right)+\hat{k}\left(x^{2}-y\right) \\ \text { Now, } \vec{a} .(\vec{b} \times \vec{c})=1\left(1-y+x^{2}\right)+0\left(x+x^{2}-x y-y\right) \\ -1\left(x^{2}-y\right) \\ =1-y+x^{2}-x^{2}+y & \end{array}$
$=1$ which shows that $\vec{a} .(\vec{b} \times \vec{c})$ does not depend
on $x$ and $y$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.