MHT CET · Maths · Vector Algebra
\(\vec{a}=\hat{i}+\hat{j}+\hat{k}\), and \(\vec{b}=\hat{i}-\hat{j}+\widehat{k}\) and \(\vec{c}=\hat{i}-\hat{j}-\hat{k}\) are three vectors then vector \(\overline{\mathrm{r}}\) in the plane of \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{b}}\), whose projection on \(\overline{\mathrm{c}}\) is \(\frac{1}{\sqrt{3}}\), is given by
- A \((2 \hat{\mathrm{i}}-1) \hat{\mathrm{i}}-\hat{\mathrm{j}}+(2 \mathrm{t}+1) \widehat{\mathrm{k}}, \forall \mathrm{t} \in \mathrm{R}\)
- B \((2 \mathrm{t}+1) \hat{\mathrm{i}}-\hat{\mathrm{j}}+(2 \mathrm{t}+1) \widehat{\mathrm{k}}, \forall \mathrm{t} \in \mathrm{R}\)
- C \((2 t-1) \hat{i}-\hat{j}+(2 t-1) \widehat{k}, \forall t \in R\)
- D \((2 t+1) \hat{i}-\hat{j}+(2 t-1) \widehat{k}, \forall t \in R\)
Answer & Solution
Correct Answer
(B) \((2 \mathrm{t}+1) \hat{\mathrm{i}}-\hat{\mathrm{j}}+(2 \mathrm{t}+1) \widehat{\mathrm{k}}, \forall \mathrm{t} \in \mathrm{R}\)
Step-by-step Solution
Detailed explanation
\(\vec{r}=t \vec{a}+u \vec{b}\) [as vector lies in the plane of \(\vec{a}\) and \(\vec{b}\) where \(t\) and \(u\) are scalars]
\(\begin{aligned} & \Rightarrow \vec{r}=t(\hat{i}+\hat{j}+\hat{k})+u(\hat{i}-\hat{j}+\hat{k}) \\ & =(t+u) \hat{i}+(t-u) \hat{j}+(t+u) \hat{k} \ldots(i) \\ & \frac{\vec{r} \cdot \vec{c}}{|\vec{c}|}=\frac{1}{\sqrt{3}} \Rightarrow \frac{(t+u)-(t-u)-(t+u)}{\sqrt{3}}=\frac{1}{\sqrt{3}} \\ & \Rightarrow-(t-u)=1 \Rightarrow u=t+1 \ldots \ldots . .(i i)\end{aligned}\)
from (i) and (ii)
\(\vec{r}=(2 t+1) \hat{i}-\hat{j}+(2 t+1) \widehat{k}\)
\(\begin{aligned} & \Rightarrow \vec{r}=t(\hat{i}+\hat{j}+\hat{k})+u(\hat{i}-\hat{j}+\hat{k}) \\ & =(t+u) \hat{i}+(t-u) \hat{j}+(t+u) \hat{k} \ldots(i) \\ & \frac{\vec{r} \cdot \vec{c}}{|\vec{c}|}=\frac{1}{\sqrt{3}} \Rightarrow \frac{(t+u)-(t-u)-(t+u)}{\sqrt{3}}=\frac{1}{\sqrt{3}} \\ & \Rightarrow-(t-u)=1 \Rightarrow u=t+1 \ldots \ldots . .(i i)\end{aligned}\)
from (i) and (ii)
\(\vec{r}=(2 t+1) \hat{i}-\hat{j}+(2 t+1) \widehat{k}\)
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