If a-b-c and b-c-a- then

The correct option is D |b|=1, |c|=|a| We have, a×b=c ⇒ca and b and b×c=a ⇒ab and c ⇒a,b,c are mutually perpendicula ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Addition of Vectors

If a×b=c an...

Question

If $\to a \times \to b = \to c$ and $\to b \times \to c = \to a,$ then

∣ ∣ \to a ∣ ∣ = 1, ∣ ∣ ∣ \to b ∣ ∣ ∣ = ∣ ∣ \to c ∣ ∣

No worries! We‘ve got your back. Try BYJU‘S free classes today!

∣ ∣ \to c ∣ ∣ = 1, ∣ ∣ \to a ∣ ∣ = 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

∣ ∣ ∣ \to b ∣ ∣ ∣ = 2, ∣ ∣ ∣ \to b ∣ ∣ ∣ = 2 ∣ ∣ \to a ∣ ∣

No worries! We‘ve got your back. Try BYJU‘S free classes today!

∣ ∣ ∣ \to b ∣ ∣ ∣ = 1, ∣ ∣ \to c ∣ ∣ = ∣ ∣ \to a ∣ ∣

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $∣ ∣ ∣ \to b ∣ ∣ ∣ = 1, ∣ ∣ \to c ∣ ∣ = ∣ ∣ \to a ∣ ∣$
We have, $\to a \times \to b = \to c$
$\Rightarrow \to c ⊥ \to a$ and $\to b$ and $\to b \times \to c = \to a$
$\Rightarrow \to a ⊥ \to b$ and $\to c$
$\Rightarrow \to a, \to b, \to c$ are mutually perpendicular
Again $\to a \times \to b = \to c$
$∣ ∣ ∣ \to a \times \to b ∣ ∣ ∣ = ∣ ∣ \to c ∣ ∣$
$\Rightarrow ∣ ∣ \to a ∣ ∣ ∣ ∣ ∣ \to b ∣ ∣ ∣ sin 90^{0} = ∣ ∣ \to c ∣ ∣$
$\Rightarrow ∣ ∣ \to a ∣ ∣ ∣ ∣ ∣ \to b ∣ ∣ ∣ = ∣ ∣ \to c ∣ ∣$ ...................... $(i)$
also, $\to b \times \to c = ∣ ∣ \to a ∣ ∣$
$∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣ sin 90^{0} = ∣ ∣ \to a ∣ ∣$
$\Rightarrow ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣ = ∣ ∣ \to a ∣ ∣$ ............ $(i i)$
From Eqs. $(i)$ and $(i i)$ , we get ${∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} ∣ ∣ \to c ∣ ∣ = ∣ ∣ \to c ∣ ∣$
$∴ {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} = 1$ $(∴ ∣ ∣ \to c ∣ ∣ \neq 0)$ or
$∣ ∣ ∣ \to b ∣ ∣ ∣ = 1$
$\Rightarrow ∣ ∣ \to a ∣ ∣ = ∣ ∣ \to c ∣ ∣$

Suggest Corrections

Similar questions

Q. If

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 and a b c = 2, then a b^{2} c^{2} + a^{2} b c^{2} + a^{2} b^{2} c =____________.

Q. if

a + b + c = 6

and

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{3}{2}

, then find

\frac{a}{b} + \frac{a}{c} + \frac{b}{a} + \frac{b}{c} + \frac{c}{a} + \frac{c}{b}

Q. If x = 1 + a², y = 1 + b², z = 1 + c² and (a + b + c)² = 0, then ab + bc + ca =

If $a b + b c + c a = 0$ , then the value of $\frac{1}{a^{2} - b c} + \frac{1}{b^{2} - c a} + \frac{1}{c^{2} - a b}$ will be:

A. $- 1$
B. $a + b + c$
C. $a b c$
D. $0$

The solution set x= $\frac{(b_{1} c_{2} - b_{2} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$ and y= $\frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$ to the pair of linear equations given below can be written as:

$a_{1}$ x + $b_{1}$ y + $c_{1}$ =0 and $a_{2}$ x + $b_{2}$ y + $c_{2}$ =0 .

Join BYJU'S Learning Program

If →a×→b=→c and →b×→c=→a, then

If $\to a \times \to b = \to c$ and $\to b \times \to c = \to a,$ then