Let PQR be a triangle- Let a-QR-b- RP and c-PQ- If -a-12- -b-4-3 and b-c-24- then which of the following is are true-

As PQR is a triangle, So, a+b+c=0 ⇒ |b+c|^2=| a|^2 ⇒ nbsp; |b|^2+|c|^2+2 b·c=|a|^2 ⇒|c|^2=48 Option (1): |c|^22 |a|=24 12=12 Option nbsp;(1) is correct. O ...

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Byju's Answer

Standard IX

Mathematics

The Mid-Point Theorem

Let PQR be a ...

Question

Let $△ P Q R$ be a triangle. Let $\to a = - - \to Q R, \to b = - - \to R P$ and $\to c = - - \to P Q .$ If $| \to a | = 12, | \to b | = 4 \sqrt{3}$ and $\to b . \to c = 24,$ then which of the following is (are) true?

\frac{| \to c |^{2}}{2} - | \to a | = 12

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\frac{| \to c |^{2}}{2} + | \to a | = 30

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| \to a \times \to b + \to c \times \to a | = 48 \sqrt{3}

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\to a . \to b = - 72

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Solution

The correct option is D $\to a . \to b = - 72$
As $P Q R$ is a triangle,
$So, \to a + \to b + \to c = 0$
$\Rightarrow | \to b + \to c |^{2} = | - \to a |^{2} \Rightarrow | \to b |^{2} + | \to c |^{2} + 2 \to b \cdot \to c = | \to a |^{2} \Rightarrow | \to c |^{2} = 48$

Option $(1)$ :
$\frac{| \to c |^{2}}{2} - | \to a | = 24 - 12 = 12$
Option $(1)$ is correct.

Option $(2)$ :
$\frac{| \to c |^{2}}{2} + | \to a | = 24 + 12 = 36$
Option $(2)$ is not correct.

$Now, \to a + \to b + \to c = 0 \Rightarrow \to a . \to b + \to b . \to b + \to c . \to b = 0 \Rightarrow \to a . \to b = - 48 - 24 = - 72$
Option $(4)$ is correct.

$\to a . \to b = - 72$
$\Rightarrow | \to a | | \to b | cos θ = - 72$
$\Rightarrow cos θ = \frac{- \sqrt{3}}{2}$
$\Rightarrow θ = \frac{5 π}{6}$

$\to a + \to b + \to c = 0$
$\Rightarrow \to a \times \to a + \to a \times \to b + \to a \times \to c = 0 \Rightarrow \to a \times \to b = - (\to a \times \to c) = \to c \times \to a$
$∴ | \to a \times \to b + \to c \times \to a | = 2 | \to a \times \to b | = 48 \sqrt{3}$
Option $(3)$ is correct.

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Similar questions

Let $Δ P Q R$ be a triangle. Let a = QR, b = RP and c = PQ. If $| a | = 12, | b | = 4 \sqrt{3}$ and $b . c = 24,$ then which of the following is/are true ?