Question

1. If a1 = λ1i + µ1j + ν1k, a2 = λ2i + µ2j + ν2k, a3 = λ3i + µ3j + ν3k, where {i,j, k} is a standard basis, show
that
(a)
a1 · (a2 × a3) =






λ1 µ1 ν1
λ2 µ2 ν2
λ3 µ3 ν3






.
(b) Deduce that a2 · (a3 × a1), due to cyclic rotation of the vectors in a triple scalar product leaves the
value of the product unchanged.
(c) If r(t) = (3t
2 − 4)i + t
3
j + (t + 3)k, where {i,j, k} is a constant standard basis, find r˙ and ¨r. Deduce
the time derivative of r × r˙.

Answer 1

$$a_1\cdot a_2\times a_3=\begin{vmatrix} \lambda_1& \mu_1 & \nu_1 \\ \lambda_2& \mu_2 & \nu_2 \\ \lambda_3& \mu_3 & \nu_3 \end{vmatrix}$$

$$a_2\times a_3=\begin{vmatrix} i& j & k \\ \lambda_2& \mu_2 & \nu_2 \\ \lambda_3& \mu_3 & \nu_3 \end{vmatrix}=$$

$$=i\begin{vmatrix} \mu_2 & \nu_2 \\ \mu_3 & \nu_3 \end{vmatrix}-j\begin{vmatrix} \lambda_2 & \nu_2 \\ \lambda_3 & \nu_3 \end{vmatrix}+k\begin{vmatrix} \lambda_2 & \mu_2 \\ \lambda_3 & \mu_3 \end{vmatrix}$$

$$a_1\cdot a_2\times a_3=\lambda_1\begin{vmatrix} \mu_2 & \nu_2 \\ \mu_3 & \nu_3 \end{vmatrix}-\mu_1\begin{vmatrix} \lambda_2 & \nu_2 \\ \lambda_3 & \nu_3 \end{vmatrix}+\nu_1\begin{vmatrix} \lambda_2 & \mu_2 \\ \lambda_3 & \mu_3 \end{vmatrix}=$$

$$=\begin{vmatrix} \lambda_1& \mu_1 & \nu_1 \\ \lambda_2& \mu_2 & \nu_2 \\ \lambda_3& \mu_3 & \nu_3 \end{vmatrix}$$

(b)

Switching Property:

The interchange of any two rows (or columns) of the determinant changes its sign.

$$a_2\cdot a_3\times a_1=\begin{vmatrix} \lambda_2& \mu_2 & \nu_2 \\ \lambda_3& \mu_3 & \nu_3 \\ \lambda_1& \mu_1 & \nu_1 \end{vmatrix}=$$

$$=-\begin{vmatrix} \lambda_2& \mu_2 & \nu_2 \\ \lambda_1& \mu_1 & \nu_1 \\ \lambda_3& \mu_3 & \nu_3 \end{vmatrix}=\begin{vmatrix} \lambda_1& \mu_1 & \nu_1 \\ \lambda_2& \mu_2 & \nu_2 \\ \lambda_3& \mu_3 & \nu_3 \end{vmatrix}=$$

$$=a_1\cdot a_2\times a_3$$

(c)

$$r(t)=(3t^2-4)i+t^3j+(t+3)k$$

$$r'(t)=6ti+3t^2j+k$$

$$r''(t)=6i+6tj+0k$$

$$r\times r'=\begin{vmatrix} i& j & k \\ 3t^2-4& t^3 & t+2 \\ 6t& 3t^2 & 1 \end{vmatrix}=$$

$$=i\begin{vmatrix} t^3 & t+2\\ 3t^2 & 1 \end{vmatrix}-j\begin{vmatrix} 3t^2-4 & t+2 \\ 6t & 1 \end{vmatrix}+k\begin{vmatrix} 3t^2-4 & t^3\\ 6t & 3t^2 \end{vmatrix}=$$

$$=(t^3-3t^3-6t^2)i-(3t^2-4-6t^2-12t)j+$$

$$+(9t^4-12t^2-6t^4)k=$$

$$=(-2t^3-6t^2)+(3t^2+12t+4)j+(3t^4-12t^2)k$$

$$(r\times r')'=(-6t^2-12t)i+(6t+12)j+(12t^3-24t)k$$