If 4i + 7j + 8k,2i + 3j + 4k and 2i + 5j + 7k | vedclass

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(A) Let the position vectors of vertices $A$,$B$,and $C$ be $\vec{a} = 4i + 7j + 8k$,$\vec{b} = 2i + 3j + 4k$,and $\vec{c} = 2i + 5j + 7k$ respectivel

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$\frac{1}{3}(6i + 13j + 18k)$

Vedclass · Answer

(A) Let the position vectors of vertices $A$,$B$,and $C$ be $\vec{a} = 4i + 7j + 8k$,$\vec{b} = 2i + 3j + 4k$,and $\vec{c} = 2i + 5j + 7k$ respectively.According to the Angle Bisector Theorem,the bisector of angle $A$ divides the opposite side $BC$ in the ratio $AB:AC$.First,we calculate the lengths of sides $AB$ and $AC$:$\vec{AB} = \vec{b} - \vec{a} = (2-4)i + (3-7)j + (4-8)k = -2i - 4j - 4k$.$|\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.$\vec{AC} = \vec{c} - \vec{a} = (2-4)i + (5-7)j + (7-8)k = -2i - 2j - k$.$|\vec{AC}| = \sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.Let $D$ be the point where the bisector of angle $A$ meets $BC$. The ratio $BD:DC = |\vec{AB}|:|\vec{AC}| = 6:3 = 2:1$.Using the section formula,the position vector of $D$ is given by $\vec{d} = \frac{1(\vec{b}) + 2(\vec{c})}{1+2} = \frac{(2i + 3j + 4k) + 2(2i + 5j + 7k)}{3}$.$\vec{d} = \frac{2i + 3j + 4k + 4i + 10j + 14k}{3} = \frac{6i + 13j + 18k}{3} = \frac{1}{3}(6i + 13j + 18k)$.

If $4i + 7j + 8k$,$2i + 3j + 4k$ and $2i + 5j + 7k$ are the position vectors of the vertices $A$,$B$ and $C$ respectively of triangle $ABC$. The position vector of the point where the bisector of angle $A$ meets $BC$ is

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