The scalar triple product of three non-zero vectors is zero if, and only if, the three vectors are coplanar.

Scalar triple product

For a given set of three vectors , * *, and * *, the scalar (Ã— * *) â‹… is called a scalar triple product of , , .

Remark

Note

Given any three vectors , , and c the following are scalar triple products:

Geometrically, the absolute value of the scalar triple product ( Ã— ) .. is the volume of the parallelepiped formed by using the three vectors , and * *as co-terminus edges. Indeed, the magnitude of the vector (* *Ã— ) is the area of the parallelogram formed by using and * *; and the direction of the vector (* *Ã— ) is perpendicular to the plane parallel to both and .

Therefore, | (* *Ã— ) â‹… * *| is | * *Ã— * *|| * *|| cos*Î¸ *| , where *Î¸* is the angle between * *Ã— and . From Fig. 6.17, we observe that | * *| | cos*Î¸ *| is the height of the parallelepiped formed by using the three vectors as adjacent vectors. Thus, | (* *Ã— * *) â‹… * *| is the volume of the parallelepiped.

The following theorem is useful for computing scalar triple products.

Theorem 6.1

*Proof*

By definition, we have

which completes the proof of the theorem.

Theorem 6.2

For any three vectors *,* and , (* *Ã— * *) â‹… * *= * *â‹… (* *Ã— ) .

Proof

Hence the theorem is proved.

Note

By Theorem 6.2, it follows that, in a scalar triple product, dot and cross can be interchanged without altering the order of occurrences of the vectors, by placing the parentheses in such a way that dot lies outside the parentheses, and cross lies between the vectors inside the parentheses. For instance, we have

Notation

For any three vectors , and * *, the scalar triple product (* *Ã— ) â‹… is denoted by [* *Ã— , ] . [* *Ã— , * *] is read as box *a*, *b*, *c *. For this reason and also because the absolute value of a scalar triple product represents the volume of a box (rectangular parallelepiped),a scalar triple product is also called a box product.

Note

(1)

In other words, [, , ] = [, , ] = [, , ] ; that is, if the three vectors are permuted in the same cyclic order, the value of the scalar triple product remains the same.

(2) If any two vectors are interchanged in their position in a scalar triple product, then the value of the scalar triple product is (-1) times the original value. More explicitly,

Theorem 6.3

The scalar triple product preserves addition and scalar multiplication. That is,

Proof

Using the properties of scalar product and vector product, we get

Using the first statement of this result, we get the following.

Similarly, the remaining equalities are proved.

We have studied about coplanar vectors in XI standard as three nonzero vectors of which, one can be expressed as a linear combination of the other two. Now we use scalar triple product for the characterisation of coplanar vectors.

Theorem 6.4

The scalar triple product of three non-zero vectors is zero if, and only if, the three vectors are coplanar.

Proof

Let , , * *be any three non-zero vectors. Then,

* *Ã— * *â‹… * *= 0 â‡” * *is perpendicular to * *Ã—

â‡” * *lies in the plane which is parallel to both * *and

, , * *are coplanar.

Three vectors , , * *are coplanar if, and only if, there exist scalars *r*, *s*, *t *âˆˆ R such that atleast one of them is non-zero and *r*+ *s** *+ *t** *= .

Proof

Applying the distributive law of cross product and using

By theorem 6.6, if ,, are non-coplanar and

then the three vectors are also non-coplanar.

If * *= âˆ’3*i*Ë†âˆ’ Ë†*j *+ 5Ë†*k*, * *= Ë†*i *âˆ’ 2 Ë†*j *+ *k*, * *= 4 Ë†*j *âˆ’ 5Ë†*k *, find * *â‹… (* *Ã— * *) .

By the defination of scalar triple product of three vectors,

, the volume of the parallelepiped is | âˆ’7 | 7 = cubic units.

Find the volume of the parallelepiped whose coterminus edges are given by the vectors 2*i*Ë† âˆ’ 3 Ë†*j *+ 4*k*Ë†, *i*Ë† + 2 Ë†*j *âˆ’ *k*Ë† and 3*i*Ë† âˆ’ Ë†*j *+ 2*k*Ë† .

We know that the volume of the parallelepiped whose coterminus edges are , , c is given by | [, , * *] | . Here, * *= 2*i*Ë†* *âˆ’ 3Ë† *j *+ 4Ë†*k*, * *= Ë†*i *+ 2 Ë†*j *âˆ’ Ë†*k*, * *= 3Ë†*i *âˆ’ Ë†*j *+ 2Ë†*k*.

Example 6.14

Show that the vectors *i*Ë† + 2 Ë†*j *âˆ’ 3*k*Ë†, 2*i*Ë† âˆ’ Ë†*j *+ 2*k*Ë† and 3*i*Ë† + Ë†*j *âˆ’ *k*Ë† are coplanar.

Solution

Therefore, the three given vectors are coplanar.

If 2*i*Ë† âˆ’ Ë†*j *+ 3*k*Ë†, 3*i*Ë† + 2 Ë†*j *+ *k*Ë†, *i*Ë† + *m*Ë†*j *+ 4*k*Ë† are coplanar, find the value of *m *.

Since the given three vectors are coplanar, we have = 0 â‡’ *m* = -3.

Show that the four points (6, âˆ’7, 0), (16, âˆ’19, âˆ’4), (0, 3, âˆ’6), (2, âˆ’5,10) lie on a same plane.

Let *A *= (6, âˆ’7, 0), *B *= (16, âˆ’19, âˆ’4), *C *= (0, 3, âˆ’6), *D *= (2, âˆ’5,10) . To show that the four points *A*, *B*, *C*, *D *lie on a plane, we have to prove that the three vectors ,, * *are coplanar.

Therefore, the three vectors ,, , are coplanar and hence the four points A, B, C, and D lie on a plane.

Example 6.17

If the vectors , , are coplanar, then prove that the vectors are also coplanar.

Since the vectors , , are coplanar, we have [, , ] = 0. Using the properties of the scalar triple product, we get

Hence the vectors are coplanar.

Example 6.18

If , , are three vectors, prove that .

Solution

Using theorem 6.6, we get

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12th Mathematics : UNIT 6 : Applications of Vector Algebra : Scalar triple product | Definition, Properties, Theorem, Proof, Solved Example Problems, Solution

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