Search any question & find its solution
Question:
Answered & Verified by Expert
If $\hat{\mathbf{a}}, \mathbf{b}$ and $\hat{\boldsymbol{c}}$ are non-coplanar vectors and if $\mathbf{d}$ is such that $\hat{\mathbf{d}}=\frac{1}{x}(\hat{\mathbf{a}}+\hat{\mathbf{b}}+\hat{\mathbf{c}})$ and $\hat{\mathbf{d}}=\frac{1}{y}(\hat{\mathbf{b}}+\hat{\mathbf{c}}+\hat{\mathbf{d}})$ where $x$ and $y$ are non-zero real numbers, then $\frac{1}{x y}(\hat{\mathbf{a}}+\hat{\mathbf{b}}+\hat{\mathbf{c}}+\hat{\mathbf{d}})$ equals to
Options:
Solution:
1156 Upvotes
Verified Answer
The correct answer is:
$0$
Given, $\mathbf{d}=\frac{1}{x}(\mathbf{a}+\mathbf{b}+\mathbf{c})$
$$
\begin{aligned}
& \text { and } \quad \mathbf{d}=\frac{1}{y}(\mathbf{b}+\mathbf{c}+\mathbf{d}) \\
& \therefore \quad \mathrm{a}+\mathrm{b}+\mathrm{c}-\mathrm{xd}=0 \\
& \text { and } \quad \mathbf{b}+\mathbf{c}+\mathbf{d}-\mathrm{yd}=0 \\
& \Rightarrow \quad \mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=0 \\
& \therefore \frac{1}{x y}(\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d})=\frac{1}{x y}(0)=0 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \text { and } \quad \mathbf{d}=\frac{1}{y}(\mathbf{b}+\mathbf{c}+\mathbf{d}) \\
& \therefore \quad \mathrm{a}+\mathrm{b}+\mathrm{c}-\mathrm{xd}=0 \\
& \text { and } \quad \mathbf{b}+\mathbf{c}+\mathbf{d}-\mathrm{yd}=0 \\
& \Rightarrow \quad \mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=0 \\
& \therefore \frac{1}{x y}(\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d})=\frac{1}{x y}(0)=0 \\
&
\end{aligned}
$$
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.