Given that A- - B- - C- and that C- is perpendicular to A- - Furtherif -A- - - - C- - -then what is the angle between A- and B- -

The correct option is C 3 π/ 4 radianLet A,B,C be vectors.Given, A+B=C A.C=0 |A|=|C| |C|^2=|A+B|^2=|A|^2+|B|^2+2|A||B|cosθ |B|+2|A ...

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Byju's Answer

Standard XII

Physics

Comparing Vectors

Given that ...

Question

Given that $\to A + \to B = \to C$ and that $\to C$ is perpendicular to $\to A$ . Furtherif $| \to A | = | \to C |$ ,then what is the angle between $\to A$ and $\to B$ ?

\frac{π}{4}

radian

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\frac{π}{2}

radian

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\frac{3 π}{4}

radian

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π

radian

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Solution

The correct option is C $\frac{3 π}{4}$ radian
Let A,B,C be vectors.

Given,

$A + B = C$

$A . C = 0$

$| A | = | C |$

$| C |^{2} = | A + B |^{2} = | A |^{2} + | B |^{2} + 2 | A | | B | cos θ$

$| B | + 2 | A | cos θ = 0$

$A . (A + B) = A . C = 0$

$| A |^{2} + A . B = 0$

$| A | + | B | cos θ = 0$

$cos θ = \frac{- | B |}{2 | A |} = \frac{- | A |}{| B |}$

$| B | = \sqrt{2} | A |$

$cos θ = \frac{- 1}{\sqrt{2}}$

$θ = \frac{3 π}{4}$

Option $C$ is the correct answer

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Similar questions

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Given that →A+→B=→C and that →C is perpendicular to →A . Furtherif |→A|=|→C| ,then what is the angle between →A and →B ?

The correct option is C 3π4 radianLet A,B,C be vectors.Given,A+B=CA.C=0|A|=|C||C|2=|A+B|2=|A|2+|B|2+2|A||B|cosθ|B|+2|A|cosθ=0A.(A+B)=A.C=0|A|2+A.B=0|A|+|B|cosθ=0cosθ=−|B|2|A|=−|A||B||B|=√2|A|cosθ=−1√2θ=3π4Option C is the correct answer

Given that $\to A + \to B = \to C$ and that $\to C$ is perpendicular to $\to A$ . Furtherif $| \to A | = | \to C |$ ,then what is the angle between $\to A$ and $\to B$ ?