Given vectors
a ⃗ = ( 9 , − 13 , 25 ) , b ⃗ = ( 3 , 7 , − 13 ) , c ⃗ = ( 17 , − 2 , 1 ) . \vec a=(9,-13,25),\quad \vec b=(3,7,-13),\quad \vec c=(17,-2,1). a = ( 9 , − 13 , 25 ) , b = ( 3 , 7 , − 13 ) , c = ( 17 , − 2 , 1 ) .
We need to find r ⃗ \vec r r
r ⃗ × a ⃗ = ( b ⃗ + c ⃗ ) × a ⃗ \vec r\times \vec a=(\vec b+\vec c)\times \vec a r × a = ( b + c ) × a r ⃗ ⋅ ( b ⃗ − c ⃗ ) = 0. \vec r\cdot(\vec b-\vec c)=0. r ⋅ ( b − c ) = 0.
Then evaluate
∣ 593 r ⃗ + 67 a ⃗ ∣ 2 593 2 . \frac{|593\vec r+67\vec a|^2}{593^2}. 59 3 2 ∣593 r + 67 a ∣ 2 .
Use the cross-product condition
From
r ⃗ × a ⃗ = ( b ⃗ + c ⃗ ) × a ⃗ , \vec r\times \vec a=(\vec b+\vec c)\times \vec a, r × a = ( b + c ) × a , ( r ⃗ − ( b ⃗ + c ⃗ ) ) × a ⃗ = 0 ⃗ . (\vec r-(\vec b+\vec c))\times \vec a=\vec 0. ( r − ( b + c )) × a = 0 .
Hence r ⃗ − ( b ⃗ + c ⃗ ) \vec r-(\vec b+\vec c) r − ( b + c ) a ⃗ \vec a a
r ⃗ = b ⃗ + c ⃗ + λ a ⃗ \vec r=\vec b+\vec c+\lambda \vec a r = b + c + λ a λ \lambda λ
Now,
b ⃗ + c ⃗ = ( 3 + 17 , 7 − 2 , − 13 + 1 ) = ( 20 , 5 , − 12 ) . \vec b+\vec c=(3+17,\,7-2,\,-13+1)=(20,5,-12). b + c = ( 3 + 17 , 7 − 2 , − 13 + 1 ) = ( 20 , 5 , − 12 ) .
Thus
r ⃗ = ( 20 , 5 , − 12 ) + λ ( 9 , − 13 , 25 ) . \vec r=(20,5,-12)+\lambda(9,-13,25). r = ( 20 , 5 , − 12 ) + λ ( 9 , − 13 , 25 ) .
Use the dot-product condition
We are given
r ⃗ ⋅ ( b ⃗ − c ⃗ ) = 0. \vec r\cdot(\vec b-\vec c)=0. r ⋅ ( b − c ) = 0.
First compute
b ⃗ − c ⃗ = ( 3 − 17 , 7 − ( − 2 ) , − 13 − 1 ) = ( − 14 , 9 , − 14 ) . \vec b-\vec c=(3-17,\,7-(-2),\,-13-1)=(-14,9,-14). b − c = ( 3 − 17 , 7 − ( − 2 ) , − 13 − 1 ) = ( − 14 , 9 , − 14 ) .
Now,
r ⃗ ⋅ ( b ⃗ − c ⃗ ) = ( ( b ⃗ + c ⃗ ) + λ a ⃗ ) ⋅ ( b ⃗ − c ⃗ ) = 0. \vec r\cdot(\vec b-\vec c)=\big((\vec b+\vec c)+\lambda\vec a\big)\cdot(\vec b-\vec c)=0. r ⋅ ( b − c ) = ( ( b + c ) + λ a ) ⋅ ( b − c ) = 0.
So,
( b ⃗ + c ⃗ ) ⋅ ( b ⃗ − c ⃗ ) + λ a ⃗ ⋅ ( b ⃗ − c ⃗ ) = 0. (\vec b+\vec c)\cdot(\vec b-\vec c)+\lambda\,\vec a\cdot(\vec b-\vec c)=0. ( b + c ) ⋅ ( b − c ) + λ a ⋅ ( b − c ) = 0.
Compute each term:
( b ⃗ + c ⃗ ) ⋅ ( b ⃗ − c ⃗ ) = ( 20 , 5 , − 12 ) ⋅ ( − 14 , 9 , − 14 ) (\vec b+\vec c)\cdot(\vec b-\vec c)=(20,5,-12)\cdot(-14,9,-14) ( b + c ) ⋅ ( b − c ) = ( 20 , 5 , − 12 ) ⋅ ( − 14 , 9 , − 14 ) = 20 ( − 14 ) + 5 ( 9 ) + ( − 12 ) ( − 14 ) = − 280 + 45 + 168 = − 67. =20(-14)+5(9)+(-12)(-14)=-280+45+168=-67. = 20 ( − 14 ) + 5 ( 9 ) + ( − 12 ) ( − 14 ) = − 280 + 45 + 168 = − 67.
Next,
a ⃗ ⋅ ( b ⃗ − c ⃗ ) = ( 9 , − 13 , 25 ) ⋅ ( − 14 , 9 , − 14 ) \vec a\cdot(\vec b-\vec c)=(9,-13,25)\cdot(-14,9,-14) a ⋅ ( b − c ) = ( 9 , − 13 , 25 ) ⋅ ( − 14 , 9 , − 14 ) = 9 ( − 14 ) + ( − 13 ) ( 9 ) + 25 ( − 14 ) = − 126 − 117 − 350 = − 593. =9(-14)+(-13)(9)+25(-14)=-126-117-350=-593. = 9 ( − 14 ) + ( − 13 ) ( 9 ) + 25 ( − 14 ) = − 126 − 117 − 350 = − 593.
Therefore,
− 67 + λ ( − 593 ) = 0 -67+\lambda(-593)=0 − 67 + λ ( − 593 ) = 0 − 67 − 593 λ = 0 -67-593\lambda=0 − 67 − 593 λ = 0 λ = − 67 593 . \lambda=-\frac{67}{593}. λ = − 593 67 .
Hence
r ⃗ = b ⃗ + c ⃗ − 67 593 a ⃗ . \vec r=\vec b+\vec c-\frac{67}{593}\vec a. r = b + c − 593 67 a .
Evaluate 593 r ⃗ + 67 a ⃗ 593\vec r+67\vec a 593 r + 67 a
Substitute r ⃗ \vec r r
593 r ⃗ + 67 a ⃗ = 593 ( b ⃗ + c ⃗ − 67 593 a ⃗ ) + 67 a ⃗ . 593\vec r+67\vec a=593\left(\vec b+\vec c-\frac{67}{593}\vec a\right)+67\vec a. 593 r + 67 a = 593 ( b + c − 593 67 a ) + 67 a .
Simplifying,
593 r ⃗ + 67 a ⃗ = 593 ( b ⃗ + c ⃗ ) − 67 a ⃗ + 67 a ⃗ = 593 ( b ⃗ + c ⃗ ) . 593\vec r+67\vec a=593(\vec b+\vec c)-67\vec a+67\vec a=593(\vec b+\vec c). 593 r + 67 a = 593 ( b + c ) − 67 a + 67 a = 593 ( b + c ) .
Therefore,
∣ 593 r ⃗ + 67 a ⃗ ∣ 2 593 2 = ∣ 593 ( b ⃗ + c ⃗ ) ∣ 2 593 2 = ∣ b ⃗ + c ⃗ ∣ 2 . \frac{|593\vec r+67\vec a|^2}{593^2}=\frac{|593(\vec b+\vec c)|^2}{593^2}=|\vec b+\vec c|^2. 59 3 2 ∣593 r + 67 a ∣ 2 = 59 3 2 ∣593 ( b + c ) ∣ 2 = ∣ b + c ∣ 2 .
Now,
b ⃗ + c ⃗ = ( 20 , 5 , − 12 ) . \vec b+\vec c=(20,5,-12). b + c = ( 20 , 5 , − 12 ) .
So,
∣ b ⃗ + c ⃗ ∣ 2 = 20 2 + 5 2 + ( − 12 ) 2 = 400 + 25 + 144 = 569. |\vec b+\vec c|^2=20^2+5^2+(-12)^2=400+25+144=569. ∣ b + c ∣ 2 = 2 0 2 + 5 2 + ( − 12 ) 2 = 400 + 25 + 144 = 569.
Final answer
569 \boxed{569} 569
The derived answer matches the stored correct answer.