vector triple product

**ayushdadhwal** · August 8th 2011, 11:09 AM

If vector b, c, d are non coplanar vectors then prove that the vector r=( aX b)X( c X d)+( aX c )X( dX b)+( aX d)X( bX c) must be parallel to vector a .
I try to solve this question by using vector triple product and at last i got
2[a, b, c] d +2[a, d, b] c+2[a, c,d] b
Where [a, b, c,] Shows scalar triple product and X means cross product. now what to do next .taking cross product with a it should give 0
But how?

**Sudharaka** · August 8th 2011, 06:08 PM

Originally Posted by ayushdadhwal

If vector b, c, d are non coplanar vectors then prove that the vector r=( aX b)X( c X d)+( aX c )X( dX b)+( aX d)X( bX c) must be parallel to vector a .
I try to solve this question by using vector triple product and at last i got
2[a, b, c] d +2[a, d, b] c+2[a, c,d] b
Where [a, b, c,] Shows scalar triple product and X means cross product. now what to do next .taking cross product with a it should give 0
But how?

Dear ayushdadhwal,

$\vec{r}=(\vec{a}\times\vec{b})\times(\vec{c}\times \vec{d})+(\vec{a}\times\vec{c})\times(\vec{d} \times\vec{b})+(\vec{a}\times\vec{d})\times(\vec{b }\times\vec{c})$

From the vector triple product expansion,

$\Rightarrow\vec{r}=2[\vec{a},\vec{b},\vec{c}].\vec{d} +2[\vec{a},\vec{d},\vec{b}].\vec{c}+2[\vec{a},\vec{c},\vec{d}].\vec{b}$

$\Rightarrow\vec{a}\times\vec{r}=2\left\{[\vec{a},\vec{b},\vec{c}].(\vec{a}\times\vec{d}) +[\vec{a},\vec{d},\vec{b}].(\vec{a}\times\vec{c})+[\vec{a},\vec{c},\vec{d}].(\vec{a}\times\vec{b})\right\}$ -------(1)

Also it follows that,

$\vec{r}=(\vec{a}\times\vec{b})\times(\vec{c}\times \vec{d})+(\vec{a}\times\vec{c})\times(\vec{d} \times\vec{b})+(\vec{a}\times\vec{d})\times(\vec{b }\times\vec{c})$

$\vec{a}\times\vec{r}=\vec{a}\times\left((\vec{a} \times \vec{b})\times(\vec{c} \times \vec{d})\right)+\vec{a} \times\left((\vec{a} \times \vec{c}) \times (\vec{d} \times\vec{b})\right)+\vec{a} \times \left((\vec{a} \times \vec{d})\times (\vec{b} \times \vec{c})\right)$

From the vector triple product expansion,

$\Rightarrow\vec{a}\times\vec{r}=[\vec{a},\vec{c},\vec{d}].(\vec{a}\times\vec{b})+[\vec{a},\vec{d},\vec{b}].(\vec{a}\times\vec{c})+[\vec{a},\vec{b},\vec{c}].(\vec{a}\times\vec{d})$ --------(2)

By (1) and (2) it is clear that, $\vec{a}\times\vec{r}=\vec{0}$ Therefore $\vec{a}$ is parallel to $\vec{r}$ .

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