If a- is a vector and x is a non-zero scalar- then

The correct option is B xa⃗ is a vector collinear to a⃗ By multiplying x which is a non zero scalar to a⃗ we get xa⃗ its magnitude ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Limits

If a⃗ is a ...

Question

If $\to a$ is a vector and $x$ is a non-zero scalar, then

x \to a

is a vector in the direction perpendicular to

\to a

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x \to a

is a vector collinear to

\to a

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x \to a

and

\to a

have independent directions

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x \to a

is a scalar

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Solution

The correct option is B $x \to a$ is a vector collinear to $\to a$
By multiplying $x$ which is a non zero scalar to $\to a$ we get $x \to a$ its magnitude increases by $x$ .
Now it is a different vector which makes an angle of zero degree with $\to a$ therefore it is collinear to $\to a$ .

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Similar questions

Q. If

\to a

is a vector and

x

is a non-zero scalar, then -

Q. Any vector

\to r

in the plane of two non-zero, non collinear vectors

\to a

and

\to b

can be expressed uniquely as a linear combination

x \to a + y \to b

\to a a n d \to b,

where

x \to a a n d y \to b

are the components of vector

\to r

Q. Let

\to a, \to b, \to c

be unit vectors such that

\to a ⊥ \to b

and

\to c

is equally inclined to

\to a

and

\to b

. If

\to c = x \to a + y \to b + z (\to a \times \to b)

, then

Q. If

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

and

| x \to a | = 1

, then

x

is equal to

Q. Assertion :Vectors

\to a, \to b, \to c

are coplanar. Reason:

(x \to a + y \to b) = \to 0 \Rightarrow

x = 0 and y = 0.

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If →a is a vector and x is a non-zero scalar, then

The correct option is B x→a is a vector collinear to →aBy multiplying x which is a non zero scalar to →a we get x→a its magnitude increases by x. Now it is a different vector which makes an angle of zero degree with →a therefore it is collinear to →a.

If $\to a$ is a vector and $x$ is a non-zero scalar, then

The correct option is B $x \to a$ is a vector collinear to $\to a$
By multiplying $x$ which is a non zero scalar to $\to a$ we get $x \to a$ its magnitude increases by $x$ .
Now it is a different vector which makes an angle of zero degree with $\to a$ therefore it is collinear to $\to a$ .