If r-2r-4r-k- and -r-21- then vector r is

Let nbsp;r=xî+yĵ+zk̂ ∵r·î=2r·ĵ=4r·k̂ ⇒ x=2y=4z⋯(i) also, |r|=√(21) ⇒ x^2+y^2+z^2=21 ⇒ 16z^2+4z^2+z^2=21 ⇒ z=±1 So, x=±4, y=±2 So, r=4î+2ĵ+k̂ or r= (4î+2ĵ+k̂) ...

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Standard XII

Mathematics

Test for Coplanarity

If r·î=2r·ĵ=4...

Question

If $\to r \cdot^i = 2 \to r \cdot^j = 4 \to r \cdot^k$ and $| \to r | = \sqrt{21},$ then vector $\to r$ is

4^i - 2^j -^k

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4^i + 2^j +^k

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- 4^i - 2^j -^k

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- 4^i + 2^j -^k

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Solution

The correct option is C $- 4^i - 2^j -^k$
Let $\to r = x^i + y^j + z^k$
$∵ \to r \cdot^i = 2 \to r \cdot^j = 4 \to r \cdot^k \Rightarrow x = 2 y = 4 z \dots (i)$
also, $| \to r | = \sqrt{21}$
$\Rightarrow x^{2} + y^{2} + z^{2} = 21 \Rightarrow 16 z^{2} + 4 z^{2} + z^{2} = 21 \Rightarrow z = \pm 1$
So, $x = \pm 4, y = \pm 2$
So, $\to r = 4^i + 2^j +^k$ or $\to r = - (4^i + 2^j +^k)$

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Test for Coplanarity

Standard XII Mathematics

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If →r⋅^i=2→r⋅^j=4→r⋅^k and |→r|=√21, then vector →r is

The correct option is C −4^i−2^j−^kLet →r=x^i+y^j+z^k ∵→r⋅^i=2→r⋅^j=4→r⋅^k⇒x=2y=4z⋯(i) also, |→r|=√21 ⇒x2+y2+z2=21⇒16z2+4z2+z2=21⇒z=±1 So, x=±4,y=±2 So, →r=4^i+2^j+^k or →r=−(4^i+2^j+^k)

If $\to r \cdot^i = 2 \to r \cdot^j = 4 \to r \cdot^k$ and $| \to r | = \sqrt{21},$ then vector $\to r$ is