(c) 1 ,(d) 2
Explanation:
This question is of multiple type.(more than one correct)
1) We have, two non null vectors 'a' and 'b' such that :
\begin{lgathered}|\vec{a}+\vec{b}|=|\vec{a}-2\vec{b}|\\ \\=>|\vec{a}+\vec{b}|^2=|\vec{a}-2\vec{b}|^2\\ \\=>|\vec{a}|^2+|\vec{b}|^2+2|\vec{a}||\vec{b}|cos(\theta)=|\vec{a}|^2+4|\vec{b}|^2-2*|\vec{a}||\vec{2b}|cos(\theta)\\ \\=>3|\vec{b}|^2=6|\vec{a}||\vec{b}|cos(\theta)\\ \\=>cos(\theta)=\frac{|\vec{b}|}{2|\vec{a}|}\\ \\=>\frac{1}{2} \leq \frac{|\vec{a}|}{|\vec{b}|}=\frac{1}{2cos(\theta)} <\infty\end{lgathered}
∣
a
+
b
∣=∣
−2
=>∣
2
=∣
+∣
+2∣
∣∣
∣cos(θ)=∣
+4∣
−2∗∣
2b
∣cos(θ)
=>3∣
=6∣
=>cos(θ)=
2∣
=>
1
≤
=
2cos(θ)
<∞
Since, 1>1/2 , 2>1/2.
So, Value of a/b may be 1 or 2.
Option (C) ,(D)
Hey your answer is in the given attachment
HOPE IT IS HELPFUL
❤#thank you#❤
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
(c) 1 ,(d) 2
Explanation:
This question is of multiple type.(more than one correct)
1) We have, two non null vectors 'a' and 'b' such that :
\begin{lgathered}|\vec{a}+\vec{b}|=|\vec{a}-2\vec{b}|\\ \\=>|\vec{a}+\vec{b}|^2=|\vec{a}-2\vec{b}|^2\\ \\=>|\vec{a}|^2+|\vec{b}|^2+2|\vec{a}||\vec{b}|cos(\theta)=|\vec{a}|^2+4|\vec{b}|^2-2*|\vec{a}||\vec{2b}|cos(\theta)\\ \\=>3|\vec{b}|^2=6|\vec{a}||\vec{b}|cos(\theta)\\ \\=>cos(\theta)=\frac{|\vec{b}|}{2|\vec{a}|}\\ \\=>\frac{1}{2} \leq \frac{|\vec{a}|}{|\vec{b}|}=\frac{1}{2cos(\theta)} <\infty\end{lgathered}
∣
a
+
b
∣=∣
a
−2
b
∣
=>∣
a
+
b
∣
2
=∣
a
−2
b
∣
2
=>∣
a
∣
2
+∣
b
∣
2
+2∣
a
∣∣
b
∣cos(θ)=∣
a
∣
2
+4∣
b
∣
2
−2∗∣
a
∣∣
2b
∣cos(θ)
=>3∣
b
∣
2
=6∣
a
∣∣
b
∣cos(θ)
=>cos(θ)=
2∣
a
∣
∣
b
∣
=>
2
1
≤
∣
b
∣
∣
a
∣
=
2cos(θ)
1
<∞
Since, 1>1/2 , 2>1/2.
So, Value of a/b may be 1 or 2.
Option (C) ,(D)
Hey your answer is in the given attachment
HOPE IT IS HELPFUL
❤#thank you#❤