Let ABC be a triangle of area and the vectors , and , d > 0. Then the square of the length of the largest side of the triangle ABC is __________. [2024]

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Solution

Q.
Let ABC be a triangle of area $15 \sqrt{2}$ and the vectors $\vec{A B} = \hat{i} + 2 \hat{j} - 7 \hat{k}$ , $\vec{B C} = a \hat{i} + b \hat{j} + c \hat{k}$ and $\vec{A C} = 6 \hat{i} + d \hat{j} - 2 \hat{k}$ , d > 0. Then the square of the length of the largest side of the triangle ABC is __________. [2024]

Ans.
(54)

Area of triangle, ABC = $15 \sqrt{2}$

$\frac{1}{2}$ $| \vec{A B} \times \vec{A C} |$ $= 15 \sqrt{2}$ ... (i)

Now, $\vec{A B} \times \vec{A C} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & - 7 \\ 6 & d & - 2 \end{matrix} |$

$= (7 d - 4) \hat{i} - 40 \hat{j} + (d - 12) \hat{k}$ ... (ii)

From (i) and (ii), we get

$\frac{1}{2} \sqrt{{(7 d - 4)}^{2} + 1600 + {(d - 12)}^{2}} = 15 \sqrt{2}$

$\Rightarrow 5 d^{2} - 8 d - 4 = 0$

$\Rightarrow d = \frac{- 2}{5}$ (Rejected as d > 0) or d = 2

Also, $\vec{A B} + \vec{B C} = \vec{A C}$

$\Rightarrow (\hat{i} + 2 \hat{j} - 7 \hat{k}) + (a \hat{i} + b \hat{j} + c \hat{k}) = 6 \hat{i} + d \hat{j} - 2 \hat{k}$

$\Rightarrow (a + 1) \hat{i} + (2 + b) \hat{j} + (- 7 + c) \hat{k} = 6 \hat{i} + 2 \hat{j} - 2 \hat{k}$

$\Rightarrow a + 1 = 6 \Rightarrow a = 5$

and b + 2 = 2 $\Rightarrow$ b = 0 and c – 7 = –2 $\Rightarrow$ c = 5

Hence, $| \bar{A B} |$ $= \sqrt{54},$ $| \bar{A C} |$ $= \sqrt{44},$ $| \bar{B C} |$ $= \sqrt{50}$

Largest side has length of $\sqrt{54}$

$∴$ $| \vec{A B} |^{2}$ $= 54$ .

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