Let three vectors {{a}},{{b}} and {c} be such that {a} × {b} = {c}, {b} × {c} = {a} and |{a}| = 2. Then which one of the following is not true? [22 Jul 2021 Shift 2]

$\vec{a} \times ((\vec{b} + \vec{c}) \times (\vec{b} - \vec{c})) = \vec{0}$
Projection of $\vec{a}$ on $(\vec{b} \times \vec{c})$ is 2
$\begin{bmatrix} \vec{a} & \vec{b} & c \end{bmatrix} + \begin{bmatrix} \vec{c} & \vec{a} & \vec{b} \end{bmatrix} = 8$
$|3 \vec{a} + \vec{b} - 2 \vec{c}|^2 = 51$

Solution:

(1)

\vec{a} \times ((\vec{b} + \vec{c}) \times (\vec{b} - \vec{c})) = \vec{0}

= \vec{a}(-\vec{b} \times \vec{c} + \vec{c} \times \vec{b})

= -2(\vec{a} \times (\vec{b} \times \vec{c}))

= -2(\vec{a} \times \vec{a}) \vec{0}

(2) Projection of

\vec{a}

(\vec{b} \times \vec{c})

= \frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{|\vec{b} \times \vec{c}|} = \frac{\vec{a} \cdot \vec{a}}{|\vec{a}|} = |\vec{a}| = 2

(3)

\begin{bmatrix} \vec{a} & \vec{b} & c \end{bmatrix} + \begin{bmatrix} \vec{c} & \vec{a} & \vec{b} \end{bmatrix}

= 2 \begin{bmatrix} \vec{a} & \vec{b} & c \end{bmatrix} = \vec{2a} \cdot (\vec{b} \times \vec{c})

= 2 \vec{a} \cdot \vec{a} = 2 |\vec{a}|^2 = 8

(4)

\vec{a} \times \vec{b} = \vec{c}

and

\vec{b} \times \vec{c} = \vec{a}

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

are mutually

\perp

vectors.

\therefore |\vec{a} \times \vec{b}| = |\vec{c}| \Rightarrow |\vec{a}| |\vec{b}| = |\vec{c}|

\Rightarrow |\vec{b}| = \frac{|\vec{c}|}{2}

Also,

|\vec{b} \times \vec{c}| = |a| \Rightarrow |\vec{b}| |\vec{c}| = 2 \Rightarrow |\vec{c}| = 2 \ \& \ |\vec{b}| = 1

|\vec{a} + \vec{b} - 2 \vec{c}|^2 = (3 \vec{a} + \vec{b} - 2 \vec{c})

\cdot (3 \vec{a} + \vec{b} - 2 \vec{c})

= 9 |\vec{a}|^2 + |\vec{b}|^2 + 4 |\vec{c}|^2

= (9 \times 4) + 1 + (4 \times 4)

= 36+1+16 = 53

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