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MHTCET 11 Apr 2026 Shift 2 Paper

Question : 138 of 150

Marks: +1, -0

|\bar{a}|=4,|\bar{b}|=3

\bar{a} \cdot \bar{b}=8

\left[\begin{array}{lll}\bar{a} & \bar{a}+\bar{b} & \bar{a} \times \bar{b}\end{array}\right]=

Solution:

Concept:

The scalar triple product

\left[ \bar{a} \ \ \bar{a}+\bar{b} \ \ \bar{a}\times \bar{b} \right]

is simplified using the property that

\left[ \bar{x} \ \ \bar{y} \ \ \bar{z} \right] = \bar{x} \cdot (\bar{y} \times \bar{z})

, and terms with repeated vectors vanish.

Explanation:

Write the given scalar triple product using linearity in the second vector:

\left[ \bar{a} \ \ \bar{a}+\bar{b} \ \ \bar{a}\times \bar{b} \right] = \left[ \bar{a} \ \ \bar{a} \ \ \bar{a}\times \bar{b} \right] + \left[ \bar{a} \ \ \bar{b} \ \ \bar{a}\times \bar{b} \right]

Since

\left[ \bar{a} \ \ \bar{a} \ \ \bar{a}\times \bar{b} \right] = 0

due to two identical vectors, we get:

\left[ \bar{a} \ \ \bar{a}+\bar{b} \ \ \bar{a}\times \bar{b} \right] = \left[ \bar{a} \ \ \bar{b} \ \ \bar{a}\times \bar{b} \right]

Now evaluate the remaining triple product:

\left[ \bar{a} \ \ \bar{b} \ \ \bar{a}\times \bar{b} \right] = \bar{a} \cdot \left( \bar{b} \times (\bar{a}\times \bar{b}) \right) = (\bar{a} \times \bar{b}) \cdot (\bar{a} \times \bar{b}) = |\bar{a} \times \bar{b}|^2

Use the identity:

|\bar{a} \times \bar{b}|^2 = |\bar{a}|^2 |\bar{b}|^2 - (\bar{a} \cdot \bar{b})^2

Substitute

|\bar{a}|=4

|\bar{b}|=3

, and

\bar{a}\cdot \bar{b}=8

|\bar{a} \times \bar{b}|^2 = 4^2 \cdot 3^2 - 8^2 = 16 \cdot 9 - 64 = 144 - 64 = 80

Therefore:

\left[ \bar{a} \ \ \bar{a}+\bar{b} \ \ \bar{a}\times \bar{b} \right] = 80

Answer:

Option B:

80

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