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Question

The value of \(\left( {\overrightarrow a - \overrightarrow b } \right).\left[ {\left( {\overrightarrow b - \overrightarrow c } \right) \times \left( {\overrightarrow c - \overrightarrow a } \right)} \right]\) is:

The correct answer is

0

Evaluating the Given Vector Expression Value

We are asked to find the value of the vector expression \(\left( {\overrightarrow a - \overrightarrow b } \right).\left[ {\left( {\overrightarrow b - \overrightarrow c } \right) \times \left( {\overrightarrow c - \overrightarrow a } \right)} \right]\). This expression is a scalar triple product involving differences of vectors.

The expression can be written in the form \((\overrightarrow u) \cdot (\overrightarrow v \times \overrightarrow w)\), where \(\overrightarrow u = \overrightarrow a - \overrightarrow b\), \(\overrightarrow v = \overrightarrow b - \overrightarrow c\), and \(\overrightarrow w = \overrightarrow c - \overrightarrow a\). We can evaluate this by first calculating the cross product \(\left( {\overrightarrow b - \overrightarrow c } \right) \times \left( {\overrightarrow c - \overrightarrow a } \right)\) and then taking the dot product with \(\left( {\overrightarrow a - \overrightarrow b } \right)\).

Calculating the Cross Product

Let's expand the cross product \(\left( {\overrightarrow b - \overrightarrow c } \right) \times \left( {\overrightarrow c - \overrightarrow a } \right)\):

Using the distributive property of the cross product:

\(\left( {\overrightarrow b - \overrightarrow c } \right) \times \left( {\overrightarrow c - \overrightarrow a } \right) = \overrightarrow b \times \left( {\overrightarrow c - \overrightarrow a } \right) - \overrightarrow c \times \left( {\overrightarrow c - \overrightarrow a } \right)\)

\( = \overrightarrow b \times \overrightarrow c - \overrightarrow b \times \overrightarrow a - \left( {\overrightarrow c \times \overrightarrow c - \overrightarrow c \times \overrightarrow a } \right)\)

We know that the cross product of a vector with itself is the zero vector, i.e., \(\overrightarrow c \times \overrightarrow c = \overrightarrow 0\).

So, the expression becomes:

\( = \overrightarrow b \times \overrightarrow c - \overrightarrow b \times \overrightarrow a - \overrightarrow 0 + \overrightarrow c \times \overrightarrow a\)

\( = \overrightarrow b \times \overrightarrow c - \overrightarrow b \times \overrightarrow a + \overrightarrow c \times \overrightarrow a\)

Using the property \(\overrightarrow u \times \overrightarrow v = - \overrightarrow v \times \overrightarrow u\), we can rewrite \(\overrightarrow b \times \overrightarrow a\) as \(-\overrightarrow a \times \overrightarrow b\). Thus:

\( = \overrightarrow b \times \overrightarrow c - (-\overrightarrow a \times \overrightarrow b) + \overrightarrow c \times \overrightarrow a\)

\( = \overrightarrow b \times \overrightarrow c + \overrightarrow a \times \overrightarrow b + \overrightarrow c \times \overrightarrow a\)

Rearranging the terms in a cyclic manner (\(\overrightarrow a \times \overrightarrow b\), \(\overrightarrow b \times \overrightarrow c\), \(\overrightarrow c \times \overrightarrow a\)):

\(\left( {\overrightarrow b - \overrightarrow c } \right) \times \left( {\overrightarrow c - \overrightarrow a } \right) = \overrightarrow a \times \overrightarrow b + \overrightarrow b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a\)

Calculating the Scalar Triple Product

Now, we need to compute the dot product of \(\left( {\overrightarrow a - \overrightarrow b } \right)\) with the result from the cross product:

\(\left( {\overrightarrow a - \overrightarrow b } \right).\left[ {\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a } \right]\)

Using the distributive property of the dot product:

\( = \overrightarrow a \cdot \left( {\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a } \right) - \overrightarrow b \cdot \left( {\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a } \right)\)

\( = \overrightarrow a \cdot \left( {\overrightarrow a \times \overrightarrow b} \right) + \overrightarrow a \cdot \left( {\overrightarrow b \times \overrightarrow c} \right) + \overrightarrow a \cdot \left( {\overrightarrow c \times \overrightarrow a} \right) - \left[ \overrightarrow b \cdot \left( {\overrightarrow a \times \overrightarrow b} \right) + \overrightarrow b \cdot \left( {\overrightarrow b \times \overrightarrow c} \right) + \overrightarrow b \cdot \left( {\overrightarrow c \times \overrightarrow a} \right) \right]\)

We can use the notation for the scalar triple product \([\overrightarrow u \overrightarrow v \overrightarrow w] = \overrightarrow u \cdot (\overrightarrow v \times \overrightarrow w)\). The expression becomes:

\( = [\overrightarrow a \overrightarrow a \overrightarrow b] + [\overrightarrow a \overrightarrow b \overrightarrow c] + [\overrightarrow a \overrightarrow c \overrightarrow a] - \left[ [\overrightarrow b \overrightarrow a \overrightarrow b] + [\overrightarrow b \overrightarrow b \overrightarrow c] + [\overrightarrow b \overrightarrow c \overrightarrow a] \right]\)

A key property of the scalar triple product is that if any two vectors are identical, the value is zero. So, \([\overrightarrow a \overrightarrow a \overrightarrow b] = 0\), \([\overrightarrow a \overrightarrow c \overrightarrow a] = 0\), \([\overrightarrow b \overrightarrow a \overrightarrow b] = 0\), and \([\overrightarrow b \overrightarrow b \overrightarrow c] = 0\).

The expression simplifies to:

\( = 0 + [\overrightarrow a \overrightarrow b \overrightarrow c] + 0 - \left[ 0 + 0 + [\overrightarrow b \overrightarrow c \overrightarrow a] \right]\)

\( = [\overrightarrow a \overrightarrow b \overrightarrow c] - [\overrightarrow b \overrightarrow c \overrightarrow a]\)

Another property of the scalar triple product is that cyclic permutation of the vectors does not change the value, i.e., \([\overrightarrow a \overrightarrow b \overrightarrow c] = [\overrightarrow b \overrightarrow c \overrightarrow a] = [\overrightarrow c \overrightarrow a \overrightarrow b]\).

Using this property, \([\overrightarrow b \overrightarrow c \overrightarrow a]\) is equal to \([\overrightarrow a \overrightarrow b \overrightarrow c]\).

So, the expression becomes:

\( = [\overrightarrow a \overrightarrow b \overrightarrow c] - [\overrightarrow a \overrightarrow b \overrightarrow c]\)

\( = 0\)

Thus, the value of the given Vector Expression Value is 0.

Conclusion

The calculation shows that \(\left( {\overrightarrow a - \overrightarrow b } \right).\left[ {\left( {\overrightarrow b - \overrightarrow c } \right) \times \left( {\overrightarrow c - \overrightarrow a } \right)} \right] = 0\).

This result aligns with one of the provided options for the Vector Expression Value.

The final answer is 0.

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Important Questions from Scalar and Vector Product

  1. If \(\overrightarrow a \) and \(\overrightarrow b\) are two unit vectors inclined to x - axis at angles 30° and 120°, then \(\left| {\overrightarrow a + \overrightarrow b } \right|\) equals

  2. The work done in moving an object along a vector \(\widehat d = 3\widehat i + 2\widehat j - 5\widehat k\) (if the applied force is \(\overrightarrow F = 2\widehat i - \widehat j - \widehat k\)) is

  3. If \(\bar a\) and \(\bar b\) are unit vectors and θ is the angle between them then \(\left| {\frac{{\bar a - \bar b}}{2}} \right|\) is

  4. Two forces F̅1 = î - ĵ + k̂ and F̅2 = 4î + 2ĵ + 3k̂ act on a particle and displace it from the point (0, 1, 2) to (1, -2, 3), then the total work done is

  5. If \(\vec{A}\) and \(\vec{B}\) are two vectors, then angle between vectors \(\vec{A}+\vec{B}\) and \(\vec{A}\times\vec{B}\) is

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