In a triangle $A B C$ , $A B = 6$ and $A C = 8$ , and angle $A = 90^{o}$ . $\to A M$ and $\to B N$ are angle bisector vectors of the angles $A$ and $B$ . If the cosine of the angle between the vectors $\to A M$ and $\to B N$ is equal to $\frac{- a}{\sqrt{b}}$ where $a$ and $b$ are coprime natural numbers then find the value of $b - a$ .

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Solution

We have,
$\begin{matrix} \frac{C N}{A N} = \frac{10}{6} = \frac{5}{3} \frac{C M}{B M} = \frac{8}{6} = \frac{4}{3} P V o f N = \frac{3 (8 ¯ j) + 5 (0)}{8} P V o f M = \frac{4 (6 ¯ j) + 3 (8 ¯ j)}{7} P V o f N = 3 ¯ j P V o f M = \frac{24}{7} ¯ i + \frac{24}{7} ¯ j - - \to A M = \frac{24}{7} ¯ i + \frac{24}{7} ¯ j - - \to B N = 3 \to i - 6 \to j = - 6 \to i + 3 \to j cos θ \frac{- \frac{144}{7} + \frac{72}{7}}{\sqrt{45} \frac{\sqrt{{(24)}^{2} + {(24)}^{2}}}{7}} = \frac{- 72}{\sqrt{45} \sqrt{{(24)}^{2} \times 2}} = \frac{- 72}{\sqrt{45} \times 24 \times \sqrt{2}} = \frac{- 3}{\sqrt{2} (3) \sqrt{5}} = \frac{- 1}{\sqrt{10}} b - a = 10 - 1 = 9 \end{matrix}$

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In a triangle ABC, AB=6 and AC=8, and angle A=90o. →AM and →BN are angle bisector vectors of the angles A and B. If the cosine of the angle between the vectors →AM and →BN is equal to −a√b where a and b are coprime natural numbers then find the value of b−a.

We have,CNAN=106=53CMBM=86=43PVofN=3(8¯j)+5(0)8PVofM=4(6¯j)+3(8¯j)7PVofN=3¯jPVofM=247¯i+247¯j−−→AM=247¯i+247¯j−−→BN=3→i−6→j=−6→i+3→jcosθ−1447+727√45√(24)2+(24)27=−72√45√(24)2×2=−72√45×24×√2=−3√2(3)√5=−1√10b−a=10−1=9