Use vector addition in triangle
In any triangle,
A B → + B C → = A C → . \overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}. A B + B C = A C . Given
A B → = i ^ + 2 j ^ − 7 k ^ , \overrightarrow{AB}=\hat i+2\hat j-7\hat k, A B = i ^ + 2 j ^ − 7 k ^ , B C → = a i ^ + b j ^ + c k ^ , \overrightarrow{BC}=a\hat i+b\hat j+c\hat k, B C = a i ^ + b j ^ + c k ^ , A C → = 6 i ^ + d j ^ − 2 k ^ . \overrightarrow{AC}=6\hat i+d\hat j-2\hat k. A C = 6 i ^ + d j ^ − 2 k ^ . So,
( a , b , c ) = ( 6 , d , − 2 ) − ( 1 , 2 , − 7 ) = ( 5 , d − 2 , 5 ) . (a,b,c)= (6,d,-2)-(1,2,-7)=(5,d-2,5). ( a , b , c ) = ( 6 , d , − 2 ) − ( 1 , 2 , − 7 ) = ( 5 , d − 2 , 5 ) . Hence
B C → = 5 i ^ + ( d − 2 ) j ^ + 5 k ^ . \overrightarrow{BC}=5\hat i+(d-2)\hat j+5\hat k. B C = 5 i ^ + ( d − 2 ) j ^ + 5 k ^ .
Use area of triangle
Area of triangle formed by vectors A B → \overrightarrow{AB} A B A C → \overrightarrow{AC} A C
1 2 ∣ A B → × A C → ∣ = 15 2 . \frac12\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|=15\sqrt2. 2 1 A B × A C = 15 2 . Therefore,
∣ A B → × A C → ∣ = 30 2 . \left|\overrightarrow{AB}\times\overrightarrow{AC}\right|=30\sqrt2. A B × A C = 30 2 . Now compute
A B → × A C → = ∣ i ^ j ^ k ^ 1 2 − 7 6 d − 2 ∣ . \overrightarrow{AB}\times\overrightarrow{AC}
=
\begin{vmatrix}
\hat i & \hat j & \hat k\\
1 & 2 & -7\\
6 & d & -2
\end{vmatrix}. A B × A C = i ^ 1 6 j ^ 2 d k ^ − 7 − 2 . Expanding,
A B → × A C → = i ^ ( 2 ⋅ ( − 2 ) − ( − 7 ) d ) − j ^ ( 1 ⋅ ( − 2 ) − ( − 7 ) ⋅ 6 ) + k ^ ( 1 ⋅ d − 2 ⋅ 6 ) . \overrightarrow{AB}\times\overrightarrow{AC}
=\hat i(2\cdot(-2)-(-7)d)-\hat j(1\cdot(-2)-(-7)\cdot6)+\hat k(1\cdot d-2\cdot6). A B × A C = i ^ ( 2 ⋅ ( − 2 ) − ( − 7 ) d ) − j ^ ( 1 ⋅ ( − 2 ) − ( − 7 ) ⋅ 6 ) + k ^ ( 1 ⋅ d − 2 ⋅ 6 ) . = ( 7 d − 4 ) i ^ − 40 j ^ + ( d − 12 ) k ^ . = (7d-4)\hat i-40\hat j+(d-12)\hat k. = ( 7 d − 4 ) i ^ − 40 j ^ + ( d − 12 ) k ^ . So,
∣ ( 7 d − 4 ) i ^ − 40 j ^ + ( d − 12 ) k ^ ∣ 2 = ( 30 2 ) 2 = 1800. |(7d-4)\hat i-40\hat j+(d-12)\hat k|^2=(30\sqrt2)^2=1800. ∣ ( 7 d − 4 ) i ^ − 40 j ^ + ( d − 12 ) k ^ ∣ 2 = ( 30 2 ) 2 = 1800. Thus,
( 7 d − 4 ) 2 + 40 2 + ( d − 12 ) 2 = 1800. (7d-4)^2+40^2+(d-12)^2=1800. ( 7 d − 4 ) 2 + 4 0 2 + ( d − 12 ) 2 = 1800. ( 49 d 2 − 56 d + 16 ) + 1600 + ( d 2 − 24 d + 144 ) = 1800. (49d^2-56d+16)+1600+(d^2-24d+144)=1800. ( 49 d 2 − 56 d + 16 ) + 1600 + ( d 2 − 24 d + 144 ) = 1800. 50 d 2 − 80 d + 1760 = 1800. 50d^2-80d+1760=1800. 50 d 2 − 80 d + 1760 = 1800. 50 d 2 − 80 d − 40 = 0. 50d^2-80d-40=0. 50 d 2 − 80 d − 40 = 0. Divide by 10:
5 d 2 − 8 d − 4 = 0. 5d^2-8d-4=0. 5 d 2 − 8 d − 4 = 0. Solving,
d = 8 ± 64 + 80 10 = 8 ± 12 10 . d=\frac{8\pm\sqrt{64+80}}{10}=\frac{8\pm12}{10}. d = 10 8 ± 64 + 80 = 10 8 ± 12 . So,
d = 2 or d = − 2 5 . d=2 \quad \text{or} \quad d=-\frac25. d = 2 or d = − 5 2 . Given d > 0 d>0 d > 0
d = 2. d=2. d = 2.
Find all side vectors and their squared lengths
We already have
A B → = ( 1 , 2 , − 7 ) , A C → = ( 6 , 2 , − 2 ) . \overrightarrow{AB}=(1,2,-7),\qquad \overrightarrow{AC}=(6,2,-2). A B = ( 1 , 2 , − 7 ) , A C = ( 6 , 2 , − 2 ) . Also,
B C → = ( 5 , d − 2 , 5 ) = ( 5 , 0 , 5 ) . \overrightarrow{BC}=(5,d-2,5)=(5,0,5). B C = ( 5 , d − 2 , 5 ) = ( 5 , 0 , 5 ) . Now,
∣ A B ∣ 2 = 1 2 + 2 2 + ( − 7 ) 2 = 1 + 4 + 49 = 54 , |AB|^2=1^2+2^2+(-7)^2=1+4+49=54, ∣ A B ∣ 2 = 1 2 + 2 2 + ( − 7 ) 2 = 1 + 4 + 49 = 54 , ∣ A C ∣ 2 = 6 2 + 2 2 + ( − 2 ) 2 = 36 + 4 + 4 = 44 , |AC|^2=6^2+2^2+(-2)^2=36+4+4=44, ∣ A C ∣ 2 = 6 2 + 2 2 + ( − 2 ) 2 = 36 + 4 + 4 = 44 , ∣ B C ∣ 2 = 5 2 + 0 2 + 5 2 = 25 + 25 = 50. |BC|^2=5^2+0^2+5^2=25+25=50. ∣ B C ∣ 2 = 5 2 + 0 2 + 5 2 = 25 + 25 = 50. Therefore, the largest side is A B AB A B
54 . \boxed{54}. 54 .
Comparison with stored answer
Stored correct answer = 54 54 54
This matches the derived answer.