Vertices of a triangle are represented by vectors A- - - - 2- - 2k- B- - 2- - - - k- and C- - 3- - - - 2k- Then-

B⃗A⃗ = O⃗A⃗ O⃗B⃗ ⇒B⃗A⃗ = î 3 ĵ + 3k̂ B⃗C⃗ = O⃗C⃗ O⃗B⃗ ⇒B⃗C⃗ = î 2 ĵ + 3k̂ |B⃗A⃗| = √(19) and |B⃗C⃗| = √(14) B⃗A⃗. B⃗C⃗ = |B⃗A⃗| |B⃗C⃗| cos θ ⇒ cos θ = ...

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Byju's Answer

Standard XII

Physics

Multiplication with Vectors

Vertices of a...

Question

Vertices of a triangle are represented by vectors $\to A =^i - 2^j + 2^k, \to B = 2^i +^j -^k$ and $\to C = 3^i -^j + 2^k$ . Then,

Length of side AB is 5 units.

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Length of side BC is of length

\sqrt{14}

units.

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Angle between side AB & BC is

θ = c o s^{- 1} \sqrt{\frac{14}{19}}

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Angle between side AB and BC is

30^{\circ}

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Solution

The correct option is C Angle between side AB & BC is $θ = c o s^{- 1} \sqrt{\frac{14}{19}}$ .

$\to B A = \to O A - \to O B$
$\Rightarrow \to B A = -^i - 3^j + 3^k$
$\to B C = \to O C - \to O B$
$\Rightarrow \to B C =^i - 2^j + 3^k$
$| \to B A | = \sqrt{19}$ and $| \to B C | = \sqrt{14}$
$\to B A . \to B C = | \to B A | | \to B C | c o s θ \Rightarrow c o s θ = \frac{\to B A . \to B C}{B A . B C}$
$c o s θ = \frac{- 1 + 6 + 9}{\sqrt{14} \sqrt{19}} = \frac{\sqrt{14}}{\sqrt{19}}$

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Similar questions

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\sqrt{78}

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Vertices of a triangle are represented by vectors →A=^i−2^j+2^k,→B=2^i+^j−^k and →C=3^i−^j+2^k. Then,

The correct option is C Angle between side AB & BC is θ=cos−1√1419. →BA=→OA−→OB ⇒→BA=−^i−3^j+3^k →BC=→OC−→OB ⇒→BC=^i−2^j+3^k |→BA|=√19 and |→BC|=√14 →BA.→BC=|→BA||→BC|cosθ⇒cosθ=→BA.→BCBA.BC cosθ=−1+6+9√14√19=√14√19

Vertices of a triangle are represented by vectors $\to A =^i - 2^j + 2^k, \to B = 2^i +^j -^k$ and $\to C = 3^i -^j + 2^k$ . Then,