If r-b-c-b- and r-a-0 where a-2-3-k- b-3-k- and c-3k- then r- is equal to

We have, r⃗×b⃗=c⃗×b⃗⇒ (r⃗ c⃗)×b⃗=0 ⇒r⃗ c⃗ is parallel to b⃗. ⇒r⃗=c⃗+λb⃗, for some λ∈ R Now, r⃗.a⃗=0 ⇒ (c⃗+λb⃗).a⃗=0 ⇒ (c⃗.a⃗)+λ(b⃗.a⃗)=0 ⇒ (2+3 3)+λ (6 3 1)=0 ...

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Byju's Answer

Standard XII

Mathematics

Applications of Cross Product

If r⃗×b⃗=c⃗×b...

Question

If $\to r \times \to b = \to c \times \to b$ and $\to r . \to a = 0$ where $\to a = 2^i + 3^j -^k, \to b = 3^i -^j +^k$ and $\to c =^i +^j + 3^k$ , then $\to r$ is equal to

$\frac{1}{2} (^i +^j +^k)$

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$2 (^i +^j +^k)$

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$2 (-^i +^j +^k)$

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$\frac{1}{2} (^i -^j +^k)$

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Solution

The correct option is C
$2 (-^i +^j +^k)$

We have, $\to r \times \to b = \to c \times \to b \Rightarrow (\to r - \to c) \times \to b = 0$
$\Rightarrow \to r - \to c$ is parallel to $\to b$ .
$\Rightarrow \to r = \to c + λ \to b$ , for some $λ \in R$
Now, $\to r . \to a = 0$
$\Rightarrow (\to c + λ \to b) . \to a = 0 \Rightarrow (\to c . \to a) + λ (\to b . \to a) = 0 \Rightarrow (2 + 3 - 3) + λ (6 - 3 - 1) = 0 \Rightarrow λ = - 1$ .

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If →r×→b=→c×→b and →r.→a=0 where →a=2^i+3^j−^k,→b=3^i−^j+^k and →c=^i+^j+3^k, then →r is equal to

The correct option is C 2(−^i+^j+^k) We have, →r×→b=→c×→b⇒(→r−→c)×→b=0 ⇒→r−→c is parallel to →b. ⇒→r=→c+λ→b, for some λ∈R Now, →r.→a=0 ⇒(→c+λ→b).→a=0⇒(→c.→a)+λ(→b.→a)=0⇒(2+3−3)+λ(6−3−1)=0⇒λ=−1.

If $\to r \times \to b = \to c \times \to b$ and $\to r . \to a = 0$ where $\to a = 2^i + 3^j -^k, \to b = 3^i -^j +^k$ and $\to c =^i +^j + 3^k$ , then $\to r$ is equal to