For two vectors A- and B- A-B-C- and -A-B-C- The angle between two vectors is-

The correct option is A 0Given: #160; C⃗=A⃗+B⃗ C⃗.C⃗=(A⃗+B⃗).(A⃗+B⃗) C^2=A^2+B^2+2ABcosθ As, #160; A+B=C #160;So, cosθ=1 #160 ...

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Standard XII

Mathematics

Collinear Vectors

For two vecto...

Question

For two vectors $\to A$ and $\to B$ , $\to A + \to B = \to C$ and $| \to A | + | \to B | = | \to C |$ . The angle between two vectors is:

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Solution

The correct option is A 0
Given:
$\to C = \to A + \to B$
$\to C . \to C = (\to A + \to B) . (\to A + \to B)$
$C^{2} = A^{2} + B^{2} + 2 A B cos θ$

As, $A + B = C$
So, $cos θ = 1$ or $θ = 0$

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For two vectors →A and →B, →A+→B=→C and |→A|+|→B|=|→C|. The angle between two vectors is:

The correct option is A 0Given: →C=→A+→B→C.→C=(→A+→B).(→A+→B)C2=A2+B2+2ABcosθAs, A+B=C So, cosθ=1 or θ=0

For two vectors $\to A$ and $\to B$ , $\to A + \to B = \to C$ and $| \to A | + | \to B | = | \to C |$ . The angle between two vectors is:

The correct option is A 0
Given:
$\to C = \to A + \to B$
$\to C . \to C = (\to A + \to B) . (\to A + \to B)$
$C^{2} = A^{2} + B^{2} + 2 A B cos θ$

As, $A + B = C$
So, $cos θ = 1$ or $θ = 0$