Let a-b be the position vectors of points A and B with respect to O and -a-a-b-b the points C and D divides A internally and externally in the ratio 2-3 If OC and OD are perpendicular then

The correct option is D 9a^2=4b^2 Using section formula,(internally) na+mbm+n OC=3×a+2×b5 Using section formula,(externally) na mbn m ...

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Byju's Answer

Standard XII

Mathematics

Section Formula

Let a,b be ...

Question

Let $\to a, \to b$ be the position vectors of points $A$ and $B$ with respect to $O$ and $∣ ∣ \to a ∣ ∣ = a, ∣ ∣ ∣ \to b ∣ ∣ ∣ = b$ the points $C$ and $D$ divides $A$ internally and externally in the ratio $2 : 3$ If $- - \to O C$ and $- - \to O D$ are perpendicular then

9 a^{2} = 4 b^{2}

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4 a^{2} = 9 b^{2}

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9 a = 4 b

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4 a = 9 b

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Solution

The correct option is D $9 a^{2} = 4 b^{2}$
Using section formula,(internally) $\frac{n \to a + m \to b}{m + n}$
$- - \to O C = \frac{3 \times \to a + 2 \times \to b}{5}$
Using section formula,(externally) $\frac{n \to a - m \to b}{n - m}$
$- - \to O D = 3 \times \to a - 2 \times \to b$
$- - \to O C ⊥$ to $- - \to O D$ (given)
$\Rightarrow ⎛ ⎝ \frac{3 \times \to a + 2 \times \to b}{5} ⎞ ⎠ (3 \times \to a - 2 \times \to b) = 0$
$(3 \times \to a + 2 \times \to b) (3 \times \to a - 2 \times \to b) = 0$
$9 \times a^{2} = 4 \times b^{2}$
or $9 a^{2} = 4 b^{2}$

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Let →a,→b be the position vectors of points A and B with respect to O and ∣∣→a∣∣=a,∣∣∣→b∣∣∣=b the points C and D divides A internally and externally in the ratio 2:3 If −−→OC and −−→OD are perpendicular then

Let $\to a, \to b$ be the position vectors of points $A$ and $B$ with respect to $O$ and $∣ ∣ \to a ∣ ∣ = a, ∣ ∣ ∣ \to b ∣ ∣ ∣ = b$ the points $C$ and $D$ divides $A$ internally and externally in the ratio $2 : 3$ If $- - \to O C$ and $- - \to O D$ are perpendicular then