Question

If \(3\) vectors \(\overline{a},\overline{b},\overline{c}\) are such that \(\overline{a}\neq \overline{0}\) and \(\overline{a}\times \overline{b}=2(\overline{a}\times \overline{c})\), \(|\overline{a}|=1\), \(|\overline{c}|=1\), \(|\overline{b}|=4\), and angle between \(\overline{b}\) and \(\overline{c}\) is \(\cos^{-1}\left(\dfrac{1}{4}\right)\) and \(\overline{b}-2\overline{c}=\lambda \overline{a}\), then \(\lambda=\)

• A. \(4\)
• B. \(3\)
• C. \(2\)
• D. \(1\)

Hint:

Use \(|\vec{u}-\vec{v}|^2=|\vec{u}|^2+|\vec{v}|^2-2\vec{u}\cdot\vec{v}\) and \(\vec{b}\cdot\vec{c}=|\vec{b}||\vec{c}|\cos\theta\) in vector magnitude problems.

Answer 1

The Correct Option is A

Step 1: Simplify the given cross product relation.
Given, \[ \overline{a}\times \overline{b}=2(\overline{a}\times \overline{c}) \] This can be written as \[ \overline{a}\times \overline{b}-2\overline{a}\times \overline{c}=0. \] Using distributive property of cross product, \[ \overline{a}\times(\overline{b}-2\overline{c})=0. \]

Step 2: Use the given vector equation.
It is also given that \[ \overline{b}-2\overline{c}=\lambda \overline{a}. \] Therefore, \[ |\overline{b}-2\overline{c}|=|\lambda \overline{a}|. \] Since \[ |\overline{a}|=1, \] we get \[ |\overline{b}-2\overline{c}|=|\lambda|. \]

Step 3: Calculate \(|\overline{b}-2\overline{c}|^2\).
\[ |\overline{b}-2\overline{c}|^2 = |\overline{b}|^2+4|\overline{c}|^2-4(\overline{b}\cdot \overline{c}). \] Given, \[ |\overline{b}|=4,\qquad |\overline{c}|=1. \] The angle between \(\overline{b}\) and \(\overline{c}\) is \[ \cos^{-1}\left(\frac{1}{4}\right). \] So, \[ \cos\theta=\frac{1}{4}. \] Therefore, \[ \overline{b}\cdot \overline{c} = |\overline{b}||\overline{c}|\cos\theta \] \[ =4\cdot1\cdot\frac{1}{4} \] \[ =1. \]

Step 4: Substitute the values.
\[ |\overline{b}-2\overline{c}|^2 = 4^2+4(1)^2-4(1) \] \[ =16+4-4 \] \[ =16. \] Thus, \[ |\overline{b}-2\overline{c}|=4. \] Since \[ |\overline{b}-2\overline{c}|=|\lambda|, \] we get \[ |\lambda|=4. \] From the given options, \[ \lambda=4. \]

Step 5: Final conclusion.
Therefore, \[ \boxed{4} \]