For three vectors a - b and c if a - b - c and a - c - b- then prove that a - b and c are mutually perpendicular vectors- - b - a - and - a -1

We are given that .a⃗×b⃗=c⃗⇒a⃗⊥c⃗ and b⃗⊥c⃗ amp; a⃗×c⃗=c⃗⇒a⃗⊥b⃗ and c⃗⊥b⃗ amp; }⇒a⃗⊥b⃗⊥c⃗....(i) Now,a⃗×b⃗=c⃗⇒ |a⃗×b⃗|=|c⃗|⇒ |a⃗||b⃗|sin π/2=|c⃗|⇒ |a⃗||b ...

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Standard XII

Mathematics

Applications of Dot Product

For three vec...

Question

For three vectors $\to a, \to b and \to c i f \to a \times \to b = \to c a n d \to a \times \to c = \to b$ ,then prove that $\to a, \to b a n d \to c$ are mutually perpendicular vectors, $| \to b | = | \to a | a n d | \to a | = 1$

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Solution

We are given that
$\begin{matrix} \to a \times \to b = \to c \Rightarrow \to a ⊥ \to c a n d \to b ⊥ \to c \to a \times \to c = \to c \Rightarrow \to a ⊥ \to b a n d \to c ⊥ \to b \end{matrix}} \Rightarrow \to a ⊥ \to b ⊥ \to c . . . . (i) N o w, \to a \times \to b = \to c \Rightarrow | \to a \times \to b | = | \to c | \Rightarrow | \to a | | \to b | s i n \frac{π}{2} = | \to c | \Rightarrow | \to a | | \to b | = | \to c | . . . (i i) [b y (i), \to a ⊥ \to b A n d, \to a \times \to c = \to b \Rightarrow | \to a | | \to c | s i n \frac{π}{2} = | \to b | \Rightarrow | \to a | | \to c | = | \to b | . . . (i i i) [b y (i), \to a ⊥ \to c B y (i i) \div (i i i), w e g e t : | \to c |^{2} = | \to b |^{2} \Rightarrow | \to c | = | \to b | . Substitute | \to c | = | \to b | i n (i i) to obtain, | \to a | = 1.$

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For three vectors →a,→b and →c if →a×→b=→c and →a×→c=→b,then prove that →a,→b and →c are mutually perpendicular vectors,|→b|=|→a| and |→a|=1

For three vectors $\to a, \to b and \to c i f \to a \times \to b = \to c a n d \to a \times \to c = \to b$ ,then prove that $\to a, \to b a n d \to c$ are mutually perpendicular vectors, $| \to b | = | \to a | a n d | \to a | = 1$