The unit vector in the direction of 2 -3 -k- isA- 2 -3 k-14B- 2 -3 -k-12C- 2 -3 -k-14D- 3 -2 -k-14

The correct option is C 2î + 3ĵ+ k̂√(14)Let a=2î + 3ĵ+ k̂ So, |a|=√(2^2+3^2+1^2)=√(14) Thus, the unit vector is given as â=a|a| ⇒ â=2î + 3ĵ+ k̂√(14) ...

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Byju's Answer

Standard XII

Physics

Unit Vectors

The unit vect...

Question

The unit vector in the direction of $2^i + 3^j +^k$ is

\frac{2^i + 3^j +^k}{\sqrt{14}}

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\frac{2^i +^j + 3^k}{\sqrt{14}}

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\frac{2^i + 3^j +^k}{\sqrt{12}}

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\frac{3^i + 2^j +^k}{\sqrt{14}}

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Solution

The correct option is A $\frac{2^i + 3^j +^k}{\sqrt{14}}$
Let $\to a = 2^i + 3^j +^k$
So, $| \to a | = \sqrt{2^{2} + 3^{2} + 1^{2}} = \sqrt{14}$
Thus, the unit vector is given as
$^a = \frac{\to a}{| \to a |}$
$\Rightarrow^a = \frac{2^i + 3^j +^k}{\sqrt{14}}$

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Similar questions

Q. The unit vector which is orthogonal to the vector

3^i + 2^j + 6^k

and is coplanar with the vectors

2^i +^j +^k a n d^i -^j +^k

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a = 2^i - 6^j - 3^k, b = 4^i + 3^j -^k

Q. The unit vector which is orthogonal to the vector

3^i + 2^j + 6^k

is coplanar with vectors

2^i +^j + 6^k a n d^i -^j -^k

Q. The vector equation of the plane which is at a distance of

\frac{3}{\sqrt{14}}

from the origin and the normal from the origin is

2^i - 3^j +^k

Q. Find a unit vector in the direction of the vector

3 \to a - 2 \to b

. Where

\to a = 2^i + 3^j +^k

and

\to b = 3^i + 2^j -^k

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The unit vector in the direction of 2^i+3^j+^k is

The correct option is A 2^i+3^j+^k√14Let →a=2^i+3^j+^k So, |→a|=√22+32+12=√14 Thus, the unit vector is given as ^a=→a|→a| ⇒ ^a=2^i+3^j+^k√14

The unit vector in the direction of $2^i + 3^j +^k$ is

The correct option is A $\frac{2^i + 3^j +^k}{\sqrt{14}}$
Let $\to a = 2^i + 3^j +^k$
So, $| \to a | = \sqrt{2^{2} + 3^{2} + 1^{2}} = \sqrt{14}$
Thus, the unit vector is given as
$^a = \frac{\to a}{| \to a |}$
$\Rightarrow^a = \frac{2^i + 3^j +^k}{\sqrt{14}}$