Question

If \(\bar{a},\bar{b}\) and \(\bar{c}\) are three non-coplanar vectors and \(\lambda\) is a real number such that \([\lambda (\bar{a}+\bar{b})\lambda ^{2}\bar{b}\lambda \bar{c}]=[\bar{a}\bar{b}+\bar{c}\bar{b}]\), then there is (are) ...

• A. exactly three values of \(\lambda\)
• B. exactly two values of \(\lambda\)
• C. exactly one value of \(\lambda\)
• D. no value of \(\lambda\)

Hint:

Pull the scalar factors out of the scalar triple product and compare both sides.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
A scalar triple product is linear in each vector, so scalars can be pulled out of each position.

Step 2: Left side
\([\lambda (\bar{a}+\bar{b}),\ \lambda ^{2}\bar{b},\ \lambda \bar{c}]=\lambda ^{4}[\bar{a}+\bar{b},\ \bar{b},\ \bar{c}]=\lambda ^{4}[\bar{a}\ \bar{b}\ \bar{c}]\).

Step 3: Right side
\([\bar{a},\ \bar{b}+\bar{c},\ \bar{b}]=[\bar{a}\ \bar{c}\ \bar{b}]=-[\bar{a}\ \bar{b}\ \bar{c}]\).

Step 4: Solve
Since the vectors are non-coplanar, \([\bar{a}\ \bar{b}\ \bar{c}]\neq 0\), so \(\lambda ^{4}=-1\), which has no real solution.

Final Answer:
There is no real value of \(\lambda\). \[ \boxed{\text{(D) }\text{no value of }\lambda} \]