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Question:
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Let $\bar{a}, \bar{b}, \bar{c}$ be three unit vectors satisfying $|\bar{a}-\bar{b}|^2+|\bar{a}-\bar{c}|^2=10$. Then
Statement (I) : $|\bar{a}+2 \bar{b}|^2+|2 \bar{a}+\bar{c}|^2=2$.
Statement (II) : $|2 \bar{a}+3 \bar{b}|^2+|3 \bar{a}+2 \bar{c}|^2=10$.
Which of the above statements is (are) true?
Options:
Statement (I) : $|\bar{a}+2 \bar{b}|^2+|2 \bar{a}+\bar{c}|^2=2$.
Statement (II) : $|2 \bar{a}+3 \bar{b}|^2+|3 \bar{a}+2 \bar{c}|^2=10$.
Which of the above statements is (are) true?
Solution:
2416 Upvotes
Verified Answer
The correct answer is:
Both Statement I and Statement II are false
Given vector experssion is $|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|^2+|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}}|^2=10$
$\begin{aligned}
& |\overrightarrow{\mathrm{a}}|^2+|\overrightarrow{\mathrm{b}}|^2-2 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+|\overrightarrow{\mathrm{a}}|^2+|\overrightarrow{\mathrm{c}}|^2-2 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=10 \\
& -2 \overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})+4=10 \\
& -2 \overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=6
\end{aligned}$
From statement (I): $|\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}|^2+|2 \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}|^2=2$
Take L.H.S,
$\begin{aligned}
& \Rightarrow|\overrightarrow{\mathrm{a}}|^2+4|\overrightarrow{\mathrm{b}}|^2+4 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+4|\overrightarrow{\mathrm{a}}|^2+|\overrightarrow{\mathrm{c}}|^2+4 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}} \\
& \Rightarrow 1+4+4 \overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})+4+1 \\
& \Rightarrow 4 \times(-3)+8+2\{\text { from (i) }\} \Rightarrow-12+10=-2
\end{aligned}$
So, statement I is false.
Now, take statement (II): $|2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}|^2+|3 \overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{c}}|^2$ Take L.H.S.
$\begin{aligned}
& |2 \vec{a}+3 \vec{b}|^2+|3 \vec{a}+2 \vec{c}|^2 \\
& =4|\vec{a}|^2+9|\vec{b}|^2+\left.|12 \vec{a} \cdot \vec{b}+9| \vec{a}\right|^2+4|\vec{c}|^2+12 \vec{a} \cdot \vec{c} \\
& =4.1+9.1+12 \vec{a} \cdot \vec{b}+9.1+4.1+12 \vec{a} \cdot \vec{c} \\
& =12 \vec{a}(\vec{b}+\vec{c})+26=12 \times-2+26=2
\end{aligned}$
So statement II is also false.
Therefore, option (d) is correct.
$\begin{aligned}
& |\overrightarrow{\mathrm{a}}|^2+|\overrightarrow{\mathrm{b}}|^2-2 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+|\overrightarrow{\mathrm{a}}|^2+|\overrightarrow{\mathrm{c}}|^2-2 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=10 \\
& -2 \overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})+4=10 \\
& -2 \overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=6
\end{aligned}$
From statement (I): $|\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}|^2+|2 \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}|^2=2$
Take L.H.S,
$\begin{aligned}
& \Rightarrow|\overrightarrow{\mathrm{a}}|^2+4|\overrightarrow{\mathrm{b}}|^2+4 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+4|\overrightarrow{\mathrm{a}}|^2+|\overrightarrow{\mathrm{c}}|^2+4 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}} \\
& \Rightarrow 1+4+4 \overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})+4+1 \\
& \Rightarrow 4 \times(-3)+8+2\{\text { from (i) }\} \Rightarrow-12+10=-2
\end{aligned}$
So, statement I is false.
Now, take statement (II): $|2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}|^2+|3 \overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{c}}|^2$ Take L.H.S.
$\begin{aligned}
& |2 \vec{a}+3 \vec{b}|^2+|3 \vec{a}+2 \vec{c}|^2 \\
& =4|\vec{a}|^2+9|\vec{b}|^2+\left.|12 \vec{a} \cdot \vec{b}+9| \vec{a}\right|^2+4|\vec{c}|^2+12 \vec{a} \cdot \vec{c} \\
& =4.1+9.1+12 \vec{a} \cdot \vec{b}+9.1+4.1+12 \vec{a} \cdot \vec{c} \\
& =12 \vec{a}(\vec{b}+\vec{c})+26=12 \times-2+26=2
\end{aligned}$
So statement II is also false.
Therefore, option (d) is correct.
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