If A- B- C- are three vectors- and A-B-C-A-2-B- and B-C-0 then

The correct option is D A⃗ . B⃗ lt; 0 B⃗.C⃗=0, B⃗⊥C⃗ By vector addition rule, from given condition A⃗ + B⃗ = C⃗ nbsp;we can say that angle between B and A ...

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Byju's Answer

Standard XII

Mathematics

Angle between Two Vectors

If A⃗, B⃗, C⃗...

Question

If $\to A, \to B, \to C$ are three vectors, and $\to A$ + $\to B$ = $\to C$ , |A|=2|B| and $\to B$ . $\to C$ = $0$ then

\to A

\to C

| = |

\to A

\to B

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\to A

\to C

| =

\to B

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\to A

\to B

< 0

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\to A

\to C

may be zero

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Solution

The correct option is C $\to A$ . $\to B$ < 0

$\to B . \to C = 0, \to B ⊥ \to C$
By vector addition rule, from given condition $\to A$ + $\to B$ = $\to C$ we can say that angle between B and A is greater than $90^{\circ}$
Hence $\to A . \to B = | \to A | | \to B | c o s θ i s - v e i . e . < 0$

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Similar questions

Q. If

\to a, \to b, \to c

are any three vectors such that

(\to a + \to b) \cdot \to c = (\to a - \to b) \cdot \to c = 0

then

(\to a \times \to b) \times \to c

Q. If

\to a, \to b, \to c

are three non-zero, non-coplanar vectors and

{\to b}_{1} = \to b - \frac{\to b \cdot \to a}{| \to a |^{2}} \to a,

{\to b}_{2} = \to b + \frac{\to b \cdot \to a}{| \to a |^{2}} \to a

{\to c}_{1} = \to c - \frac{\to c \cdot \to a}{| \to a |^{2}} \to a + \frac{\to b \cdot \to c}{| \to c |^{2}} {\to b}_{1},

{\to c}_{2} = \to c - \frac{\to c \cdot \to a}{| \to a |^{2}} \to a - \frac{{\to b}_{1} \cdot \to c}{| {\to b}_{1} |^{2}} {\to b}_{1},

{\to c}_{3} = \to c - \frac{\to c \cdot \to a}{| \to c |^{2}} \to a + \frac{\to b \cdot \to c}{| \to c |^{2}} {\to b}_{1}, {\to c}_{4} = \to c - \frac{\to c \cdot \to a}{| \to c |^{2}} \to a - \frac{\to b \cdot \to c}{| \to b |^{2}} {\to b}_{1}

, then the set of orthogonal vectors is

Q. If

\to a, \to b, \to c

are three vectors and

| \to b | = | \to c |

then

(\to a + \to b) \times (\to a + \to c)} \times (\to b \times \to c) . (\to b + \to c) =

Q. For three non coplanar vectors

\to a, \to b, \to c

the relation

∣ ∣ \to a \times \to b \cdot \to c ∣ ∣ = | \to a | ∣ ∣ \to b ∣ ∣ | \to c |

holds if and only if

Q. Let

\to a, \to b

and

\to c

be three vectors such that

| \to a | = \sqrt{3}, | \to b | = 5, \to b . \to c = 10

and the angle between

\to b

and

\to c

\frac{π}{3} .

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is perpendicular to vector

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, then

| \to a \times (\to b \times \to c) |

is equal to

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If →A,→B,→C are three vectors, and →A + →B = →C, |A|=2|B| and →B.→C= 0 then

The correct option is C →A . →B < 0 →B.→C=0,→B⊥→C By vector addition rule, from given condition →A + →B = →C we can say that angle between B and A is greater than 90∘ Hence →A.→B=|→A||→B|cosθ is −ve i.e.<0

If $\to A, \to B, \to C$ are three vectors, and $\to A$ + $\to B$ = $\to C$ , |A|=2|B| and $\to B$ . $\to C$ = $0$ then

The correct option is C $\to A$ . $\to B$ < 0

$\to B . \to C = 0, \to B ⊥ \to C$
By vector addition rule, from given condition $\to A$ + $\to B$ = $\to C$ we can say that angle between B and A is greater than $90^{\circ}$
Hence $\to A . \to B = | \to A | | \to B | c o s θ i s - v e i . e . < 0$