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Chapter 14: Problem 16

If the triple scalar product \((\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}\) is equal to zero, then (i) \(a=0\), or \(b=0\), or \(c=0\) or (ii) two of the vectors are parallel, or (iii) the three vectors lie in the same plane (they are said to be coplanar). Show that the vectors $$ 2 \boldsymbol{i}-\boldsymbol{j}+\boldsymbol{k}, 3 \boldsymbol{i}-4 \boldsymbol{j}+5 \boldsymbol{k}, \boldsymbol{i}+2 \boldsymbol{j}-3 \boldsymbol{k} $$ are coplanar.

Short Answer

Expert verified
Answer: Yes, the vectors are coplanar since their triple scalar product is equal to zero.

Step by step solution

01

Find the cross product of the first two vectors.

Let the first two vectors be represented as \(\boldsymbol{a} = (2, -1, 1)\) and \(\boldsymbol{b} = (3, -4, 5)\). Calculate their cross product \(\boldsymbol{a} \times \boldsymbol{b}\) using the formula $$ \boldsymbol{a} \times \boldsymbol{b} = \begin{pmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{pmatrix} $$
02

Evaluate the determinant.

Expanding the determinant, we have: $$ \boldsymbol{a} \times \boldsymbol{b} = \hat{i}(a_2b_3 - a_3b_2) - \hat{j}(a_1b_3 - a_3b_1) + \hat{k}(a_1b_2 - a_2b_1) $$ After substituting the components of \(\boldsymbol{a}\) and \(\boldsymbol{b}\), we get: $$ \boldsymbol{a} \times \boldsymbol{b} = \hat{i}((-1)(5) - (1)(-4)) - \hat{j}((2)(5) - (1)(3)) + \hat{k}((2)(-4) - (-1)(3)) $$
03

Simplify the cross product.

Now, simplify the cross product: $$ \boldsymbol{a} \times \boldsymbol{b} = \hat{i}(5+4) - \hat{j}(10-3) + \hat{k}(-8+3) \\ \boldsymbol{a} \times \boldsymbol{b} = 9\hat{i} - 7\hat{j} - 5\hat{k} $$
04

Calculate the dot product with the third vector.

Let the third vector be represented as \(\boldsymbol{c} = (1, 2, -3)\). Calculate the dot product \((\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}\) as follows: $$ (\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c} = (9\hat{i} - 7\hat{j} - 5\hat{k}) \cdot (1\hat{i} + 2\hat{j} - 3\hat{k}) \\ = 9(1) - 7(2) - 5(-3) $$
05

Simplify the dot product.

Finally, simplify the dot product: $$ (\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c} = 9 - 14 + 15 \\ (\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c} = 0 $$ Since the triple scalar product is equal to zero, the vectors are coplanar.

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Most popular questions from this chapter

Find the volume of the parallelepiped whose edges are represented by the vectors \(12 i+j+k, 2 i\) and \(-2 \boldsymbol{j}+\boldsymbol{k}\)

Find the modulus of each of the following vectors: (a) \(\boldsymbol{r}=7 \boldsymbol{i}+3 \boldsymbol{j}\) (b) \(\boldsymbol{r}=17 \boldsymbol{i}\) (c) \(\boldsymbol{r}=2 \boldsymbol{i}-3 \boldsymbol{j}\) (d) \(r=-3 j\) (f) \(\boldsymbol{r}=a \boldsymbol{i}-b \boldsymbol{j}\)

Find the modulus of the vector $$ p=2 i-j+5 k $$

On a diagram show the arbitrary vectors \(p\) and q. Then show the following: (a) \(p+q\) (d) \(4 q\) (e) \(-2 q\) (c) \(q-p\).

State the coordinates of the point \(\mathrm{P}\) if its position vector is given as (a) \(3 \boldsymbol{i}-7 \boldsymbol{j}\), (b) \(-4 \boldsymbol{i}\), (c) \(-0.5 \boldsymbol{i}+13 \boldsymbol{j}\), (d) \(a \boldsymbol{i}+b \boldsymbol{j}\).
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