If x- is a vector in the direction of 2-2-1 of magnitude 6 and y- is a vector in the direction of 1-1-1 of magnitude -3- then -x-2 y- is equal toA- 40B- -35C- -17D- 2 -10

The correct option is D 2√(10)x=6(2î 2ĵ+ k̂)√(4+4+1) and y=√(3)(î+ĵ k̂)√(1+1+1) ⇒ nbsp;x=4î 4ĵ+ 2k̂ and y=î+ĵ k̂ ⇒x+2y=4î 4ĵ+2k̂+2î+2ĵ 2k̂ ⇒x+2y=6î 2ĵ |x+2y ...

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Byju's Answer

Standard XII

Mathematics

Addition of Vectors

If x⃗ is a ve...

Question

If $\to x$ is a vector in the direction of $(2, - 2, 1)$ of magnitude $6$ and $\to y$ is a vector in the direction of $(1, 1, - 1)$ of magnitude $\sqrt{3}$ , then $| \to x + 2 \to y |$ is equal to

40

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\sqrt{35}

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\sqrt{17}

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2 \sqrt{10}

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Solution

The correct option is D $2 \sqrt{10}$
$\to x = \frac{6 (2^i - 2^j +^k)}{\sqrt{4 + 4 + 1}}$
and $\to y = \frac{\sqrt{3} (^i +^j -^k)}{\sqrt{1 + 1 + 1}}$

$\Rightarrow \to x = 4^i - 4^j + 2^k$
and $\to y =^i +^j -^k$
$\Rightarrow \to x + 2 \to y = 4^i - 4^j + 2^k + 2^i + 2^j - 2^k$
$\Rightarrow \to x + 2 \to y = 6^i - 2^j$
$| \to x + 2 \to y | = \sqrt{36 + 4}$
$\Rightarrow | \to x + 2 \to y | = 2 \sqrt{10}$

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If →x is a vector in the direction of (2,−2,1) of magnitude 6 and →y is a vector in the direction of (1,1,−1) of magnitude √3, then |→x+2→y| is equal to

The correct option is D 2√10→x=6(2^i−2^j+^k)√4+4+1 and →y=√3(^i+^j−^k)√1+1+1 ⇒→x=4^i−4^j+2^k and →y=^i+^j−^k ⇒→x+2→y=4^i−4^j+2^k+2^i+2^j−2^k ⇒→x+2→y=6^i−2^j |→x+2→y|=√36+4 ⇒|→x+2→y|=2√10

If $\to x$ is a vector in the direction of $(2, - 2, 1)$ of magnitude $6$ and $\to y$ is a vector in the direction of $(1, 1, - 1)$ of magnitude $\sqrt{3}$ , then $| \to x + 2 \to y |$ is equal to