Find a vector v- which is co-planar with the vectors a- - -2- -k- and b- - - -2k- and is orthogonal to the vector c- -2- - -k- It is given that the projection of v- along the vector - - -k- is equal to 16- 3 -

The correct option is B 12î 6ĵ +30k̂ A vector co planar with #⃗1⃗6⃗0⃗;⃗a⃗ and #⃗1⃗6⃗0⃗;⃗b⃗ and orthogonal to #⃗1⃗6⃗0⃗;⃗c⃗ is parallel t ...

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Byju's Answer

Standard XII

Mathematics

Direction Cosines

Find a vector...

Question

Find a vector $\to v$ which is co-planar with the vectors $\to a =^i - 2^j +^k$ and $\to b =^i -^j + 2^k$ and is orthogonal to the vector $\to c = - 2^i +^j +^k$ . It is given that the projection of $\to v$ along the vector $^i -^j +^k$ is equal to $16 \sqrt{3}$ .

6^j - 12^k

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12^i - 6^j + 30^k

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9 (^i -^j +^k)

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None of these

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Solution

The correct option is B $12^i - 6^j + 30^k$
A vector co-planar with $\to a$ and $\to b$ and orthogonal to $\to c$ is parallel to the triple product
$(\to a \times \to b) \times \to c$
Given as
$(\to a \times \to b) \times \to c = (\to a . \to c) \to b - (\to b . \to c) \to a$

$\to v = α ((\to a . \to c) \to b - (\to b . \to c) \to a)$
Hence
$\to v = α [- 3 (^i -^j + 2^k) + (^i - 2^j +^k)]$
$= α [- 2^i +^j - 5^k]$

Projection of $^v$ along $^i -^j +^k$ , is
$\frac{\to v . (^i -^j +^k)}{∣ ∣ \to i - \to j + \to k ∣ ∣} = 16 \sqrt{3}$

$α [- 2^i +^j -^k] . (^i -^j +^k) = 48$

$\Rightarrow α (- 8) = 48$
$\Rightarrow α = - 6$
$\Rightarrow \to v = 12^i - 6^j + 30^k$

Hence, option $B$ .

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Find a vector →v which is co-planar with the vectors →a=^i−2^j+^k and →b=^i−^j+2^k and is orthogonal to the vector →c=−2^i+^j+^k. It is given that the projection of →v along the vector ^i−^j+^k is equal to 16√3.

Find a vector $\to v$ which is co-planar with the vectors $\to a =^i - 2^j +^k$ and $\to b =^i -^j + 2^k$ and is orthogonal to the vector $\to c = - 2^i +^j +^k$ . It is given that the projection of $\to v$ along the vector $^i -^j +^k$ is equal to $16 \sqrt{3}$ .