Name: Geometrical Interpretation of Scalar Triple Product
Uploaded: 2019-07-19
Duration: 14 min
Description: Do you know what is the geometrical meaning of the scalar triple product? And why the scalar triple product so important? Check out the video to explore.

Question

If

\vec{a}, \vec{b}, \vec{c}

are three non-zero, non-coplanar vectors, establish a linear relationship between the vectors

\vec{a} + 3 \vec{b} + 4 \vec{c}, \vec{a} - 2 \vec{b} + 3 \vec{c}, \vec{a} + 5 \vec{b} - 2 \vec{c}

and

6 \vec{a} + 14 \vec{b} + 4 \vec{c}

.

Accepted Answer

Let the four vectors in the question be written as $\vec{A} = \vec{a} + 3 \vec{b} + 4 \vec{c}, \vec{B} = \vec{a} - 2 \vec{b} + 3 \vec{c}, \vec{C} = \vec{a} + 5 \vec{b} - 2 \vec{c}, \vec{D} = 6 \vec{a} + 14 \vec{b} + 4 \vec{c}$

They can be related to each other as below:

$\vec{D} = x \vec{A} + y \vec{B} + z \vec{C}$

$6 \vec{a} + 14 \vec{b} + 4 \vec{c} = x (\vec{a} + 3 \vec{b} + 4 \vec{c}) + y (\vec{a} - 2 \vec{b} + 3 \vec{c}) + z (\vec{a} + 5 \vec{b} - 2 \vec{c})$

$6 \vec{a} + 14 \vec{b} + 4 \vec{c} = x \vec{a} + 3 x \vec{b} + 4 x \vec{c} + y \vec{a} - 2 y \vec{b} + 3 y \vec{c} + z \vec{a} + 5 z \vec{b} - 2 z \vec{c}$

$6 \vec{a} + 14 \vec{b} + 4 \vec{c} = (x + y + z) \vec{a} + (3 x - 2 y + 5 z) \vec{b} + (4 x + 3 y - 2 z) \vec{c}$

Now equating the coefficients of $\vec{a}, \vec{b}, \vec{c}$ , we get three equations as below (since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar):

$6 = (x + y + z) \Rightarrow x + y + z = 6 \dots (i)$

$14 = (3 x - 2 y + 5 z) \Rightarrow 3 x - 2 y + 5 z = 14 \dots (ii)$

$4 = (4 x + 3 y - 2 z) \Rightarrow 4 x + 3 y - 2 z = 4 \dots (iii)$

From equation $(i)$ , we get $x = 6 - y - z$ . Substituting this in equation $(ii)$ and $(iii)$ ,

$3 (6 - y - z) - 2 y + 5 z = 14$

$18 - 3 y - 3 z - 2 y + 5 z = 14$

$- 5 y + 2 z = - 4 \dots (iv)$

$4 (6 - y - z) + 3 y - 2 z = 4$

$24 - 4 y - 4 z + 3 y - 2 z = 4$

$y + 6 z = 20 \dots (v)$

Multiplying equation $(v)$ with $5$ and solving equations $(iv)$ and $(v)$ ,

$\frac{- 5 y + 2 z = - 4 5 y + 30 z = 100}{0 + 32 z = 96} z = 3$

Substituting $z = 3$ in equation $(v)$ ,

$y + 6 (3) = 20$

$y = 20 - 18$

$y = 2$

Now, $x = 6 - y - z \Rightarrow 6 - 2 - 3 \Rightarrow x = 1$ .

Therefore, $x = 1, y = 2$ and $z = 3$ .

Now the four vectors can be related linearly as:

$\vec{D} = x \vec{A} + y \vec{B} + z \vec{C}$

$6 \vec{a} + 14 \vec{b} + 4 \vec{c} = (\vec{a} + 3 \vec{b} + 4 \vec{c}) + 2 (\vec{a} - 2 \vec{b} + 3 \vec{c}) + 3 (\vec{a} + 5 \vec{b} - 2 \vec{c})$

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