Question

Let $\sf{\vec{r},\vec{a},\vec{b}$ and $\sf{\vec{c}}$ be four non-zero vectors such that $\vec{r}\cdotp\vec{a} = 0, [tex]\sf{|\vec{r}\times\vec{b}|=|\vec{r}||\vec{b}|, |\vec{r}\times\vec{c}|=|\vec{r}||\vec{c}|}}$ , then find [a b c].

(1) |a| |b| |c|
(2) -|a| |b| |c|
(3) 0
(4) None​

Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\: |\vec{r}| \ne \: 0$

$\rm :\longmapsto\: |\vec{a}| \ne \: 0$

$\rm :\longmapsto\: |\vec{b}| \ne \: 0$

$\rm :\longmapsto\: |\vec{c}| \ne \: 0$

Also, Given that,

$\rm :\longmapsto\:\vec{r} \: . \: \vec{a} = 0$

$\bf\implies \:\vec{r} \: \perp \: \vec{a} - - - (1)$

Again,

$\rm :\longmapsto\: |\vec{r} \times \vec{b}| = |\vec{r}| |\vec{b}|$

can be rewritten as

$\rm :\longmapsto\: |\vec{r} \times \vec{b}| = |\vec{r}| |\vec{b}| \times 1$

$\rm :\longmapsto\: |\vec{r} \times \vec{b}| = |\vec{r}| |\vec{b}|sin \dfrac{\pi}{2}$

$\rm :\implies\:Angle \: between \: \vec{r} \: and \: \vec{b} \: is \: \dfrac{\pi}{2}$

$\bf\implies \:\vec{r} \: \perp \: \vec{b} - - - (2)$

Again,

$\rm :\longmapsto\: |\vec{r} \times \vec{c}| = |\vec{r}| |\vec{c}|$

can be rewritten as

$\rm :\longmapsto\: |\vec{r} \times \vec{c}| = |\vec{r}| |\vec{c}| \times 1$

$\rm :\longmapsto\: |\vec{r} \times \vec{c}| = |\vec{r}| |\vec{c}|sin \dfrac{\pi}{2}$

$\rm :\implies\:Angle \: between \: \vec{r} \: and \: \vec{c} \: is \: \dfrac{\pi}{2}$

$\bf\implies \:\vec{r} \: \perp \: \vec{c} - - - (3)$

Now, from equation (1), (2) and (3), we concluded that

$\rm :\longmapsto\:\vec{a}, \: \vec{b}, \: \vec{c} \: are \: coplanar$

Using scalar triple product,

$\bf\implies \:\bigg[ \vec{a} \: \: \: \: \vec{b} \: \: \: \: \vec{c}\bigg] = 0$

Hence,

• ★ Option 3 is correct

Additional Information :-

$\boxed{ \sf{ \: \bigg[\vec{a} \: \vec{b} \: \vec{c} \bigg] = \vec{a}. \: (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}). \: \vec{c}}}$

$\boxed{ \sf{ \: \bigg[\vec{a} \: \vec{b} \: \vec{c} \bigg] = - \: \bigg[ \vec{b} \: \vec{a} \: \vec{c}\bigg]}}$

$\boxed{ \sf{ \: \bigg[\vec{a} \: \vec{b} \: \vec{c} \bigg] = 0 \: \implies \: \vec{a}, \: \vec{b}, \: \vec{c} \: are \: coplanar}}$

$\boxed{ \sf{ \: \bigg[\vec{a} \: \vec{a} \: \vec{c} \bigg] = 0 \: }}$

$\boxed{ \sf{ \: |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \: sin \theta}}$

$\boxed{ \sf{ \: \vec{a} \: . \: \vec{b} = |\vec{a}| |\vec{b}| \: cos \theta}}$