Given vectors and conditions
We have
a ⃗ = i ^ + 2 j ^ + λ k ^ , b ⃗ = 3 i ^ − 5 j ^ − λ k ^ \vec a = \hat i + 2\hat j + \lambda \hat k, \qquad \vec b = 3\hat i - 5\hat j - \lambda \hat k a = i ^ + 2 j ^ + λ k ^ , b = 3 i ^ − 5 j ^ − λ k ^ a ⃗ ⋅ c ⃗ = 7 , \vec a\cdot \vec c = 7, a ⋅ c = 7 , 2 b ⃗ ⋅ c ⃗ + 43 = 0 ⟹ b ⃗ ⋅ c ⃗ = − 43 2 , 2\vec b\cdot \vec c + 43 = 0 \implies \vec b\cdot \vec c = -\frac{43}{2}, 2 b ⋅ c + 43 = 0 ⟹ b ⋅ c = − 2 43 , a ⃗ × c ⃗ = b ⃗ × c ⃗ . \vec a \times \vec c = \vec b \times \vec c. a × c = b × c .
We need to find
∣ a ⃗ ⋅ b ⃗ ∣ . \left|\vec a\cdot \vec b\right|. a ⋅ b .
Use the cross product condition
Given
a ⃗ × c ⃗ = b ⃗ × c ⃗ , \vec a \times \vec c = \vec b \times \vec c, a × c = b × c , ( a ⃗ − b ⃗ ) × c ⃗ = 0 ⃗ . (\vec a - \vec b) \times \vec c = \vec 0. ( a − b ) × c = 0 .
This means a ⃗ − b ⃗ \vec a - \vec b a − b c ⃗ \vec c c
Now,
a ⃗ − b ⃗ = ( 1 − 3 ) i ^ + ( 2 − ( − 5 ) ) j ^ + ( λ − ( − λ ) ) k ^ \vec a - \vec b = (1-3)\hat i + (2-(-5))\hat j + (\lambda-(-\lambda))\hat k a − b = ( 1 − 3 ) i ^ + ( 2 − ( − 5 )) j ^ + ( λ − ( − λ )) k ^ = − 2 i ^ + 7 j ^ + 2 λ k ^ . = -2\hat i + 7\hat j + 2\lambda \hat k. = − 2 i ^ + 7 j ^ + 2 λ k ^ .
Hence let
c ⃗ = t ( a ⃗ − b ⃗ ) \vec c = t(\vec a - \vec b) c = t ( a − b ) t t t
Apply dot product conditions
Since
c ⃗ = t ( a ⃗ − b ⃗ ) , \vec c = t(\vec a - \vec b), c = t ( a − b ) , a ⃗ ⋅ c ⃗ = t a ⃗ ⋅ ( a ⃗ − b ⃗ ) = 7 \vec a\cdot \vec c = t\,\vec a\cdot(\vec a-\vec b)=7 a ⋅ c = t a ⋅ ( a − b ) = 7 b ⃗ ⋅ c ⃗ = t b ⃗ ⋅ ( a ⃗ − b ⃗ ) = − 43 2 . \vec b\cdot \vec c = t\,\vec b\cdot(\vec a-\vec b)=-\frac{43}{2}. b ⋅ c = t b ⋅ ( a − b ) = − 2 43 .
Now compute the needed dot products.
Let
x = a ⃗ ⋅ b ⃗ . x = \vec a\cdot \vec b. x = a ⋅ b . a ⃗ ⋅ a ⃗ = 1 2 + 2 2 + λ 2 = 5 + λ 2 , \vec a\cdot \vec a = 1^2+2^2+\lambda^2 = 5+\lambda^2, a ⋅ a = 1 2 + 2 2 + λ 2 = 5 + λ 2 , b ⃗ ⋅ b ⃗ = 3 2 + ( − 5 ) 2 + ( − λ ) 2 = 34 + λ 2 . \vec b\cdot \vec b = 3^2+(-5)^2+(-\lambda)^2 = 34+\lambda^2. b ⋅ b = 3 2 + ( − 5 ) 2 + ( − λ ) 2 = 34 + λ 2 .
Therefore,
a ⃗ ⋅ ( a ⃗ − b ⃗ ) = a ⃗ ⋅ a ⃗ − a ⃗ ⋅ b ⃗ = ( 5 + λ 2 ) − x , \vec a\cdot(\vec a-\vec b)=\vec a\cdot\vec a-\vec a\cdot\vec b=(5+\lambda^2)-x, a ⋅ ( a − b ) = a ⋅ a − a ⋅ b = ( 5 + λ 2 ) − x , b ⃗ ⋅ ( a ⃗ − b ⃗ ) = a ⃗ ⋅ b ⃗ − b ⃗ ⋅ b ⃗ = x − ( 34 + λ 2 ) . \vec b\cdot(\vec a-\vec b)=\vec a\cdot\vec b-\vec b\cdot\vec b=x-(34+\lambda^2). b ⋅ ( a − b ) = a ⋅ b − b ⋅ b = x − ( 34 + λ 2 ) .
So,
t ( ( 5 + λ 2 ) − x ) = 7 . . . ( 1 ) t\big((5+\lambda^2)-x\big)=7 \quad ...(1) t ( ( 5 + λ 2 ) − x ) = 7 ... ( 1 ) t ( x − ( 34 + λ 2 ) ) = − 43 2 . . . ( 2 ) t\big(x-(34+\lambda^2)\big)=-\frac{43}{2} \quad ...(2) t ( x − ( 34 + λ 2 ) ) = − 2 43 ... ( 2 )
Eliminate t t t λ \lambda λ
Add the bracketed expressions from (1) and (2):
( ( 5 + λ 2 ) − x ) + ( x − ( 34 + λ 2 ) ) = − 29. ((5+\lambda^2)-x) + (x-(34+\lambda^2)) = -29. (( 5 + λ 2 ) − x ) + ( x − ( 34 + λ 2 )) = − 29.
Thus the two bracketed quantities differ only by a constant sum.
From (1) and (2), divide by t t t 7 t = ( 5 + λ 2 ) − x , \frac{7}{t} = (5+\lambda^2)-x, t 7 = ( 5 + λ 2 ) − x , − 43 2 t = x − ( 34 + λ 2 ) . -\frac{43}{2t} = x-(34+\lambda^2). − 2 t 43 = x − ( 34 + λ 2 ) .
Adding,
7 t − 43 2 t = − 29. \frac{7}{t} - \frac{43}{2t} = -29. t 7 − 2 t 43 = − 29.
So,
14 − 43 2 t = − 29 \frac{14-43}{2t} = -29 2 t 14 − 43 = − 29 − 29 2 t = − 29 \frac{-29}{2t} = -29 2 t − 29 = − 29 2 t = 1 ⟹ t = 1 2 . 2t=1 \implies t=\frac12. 2 t = 1 ⟹ t = 2 1 .
Now use (1):
1 2 ( ( 5 + λ 2 ) − x ) = 7 \frac12\big((5+\lambda^2)-x\big)=7 2 1 ( ( 5 + λ 2 ) − x ) = 7 ( 5 + λ 2 ) − x = 14 (5+\lambda^2)-x=14 ( 5 + λ 2 ) − x = 14 x = λ 2 − 9. . . . ( 3 ) x=\lambda^2-9. \quad ...(3) x = λ 2 − 9. ... ( 3 )
Use (2):
1 2 ( x − ( 34 + λ 2 ) ) = − 43 2 \frac12\big(x-(34+\lambda^2)\big)=-\frac{43}{2} 2 1 ( x − ( 34 + λ 2 ) ) = − 2 43 x − ( 34 + λ 2 ) = − 43 x-(34+\lambda^2)=-43 x − ( 34 + λ 2 ) = − 43 x = λ 2 − 9 , x=\lambda^2-9, x = λ 2 − 9 ,
Use direct computation of a ⃗ ⋅ b ⃗ \vec a\cdot\vec b a ⋅ b
Now compute
a ⃗ ⋅ b ⃗ = ( 1 ) ( 3 ) + ( 2 ) ( − 5 ) + ( λ ) ( − λ ) \vec a\cdot\vec b = (1)(3) + (2)(-5) + (\lambda)(-\lambda) a ⋅ b = ( 1 ) ( 3 ) + ( 2 ) ( − 5 ) + ( λ ) ( − λ ) = 3 − 10 − λ 2 = − 7 − λ 2 . . . . ( 4 ) =3-10-\lambda^2 = -7-\lambda^2. \quad ...(4) = 3 − 10 − λ 2 = − 7 − λ 2 . ... ( 4 )
But from (3),
a ⃗ ⋅ b ⃗ = λ 2 − 9. . . . ( 5 ) \vec a\cdot\vec b = \lambda^2-9. \quad ...(5) a ⋅ b = λ 2 − 9. ... ( 5 )
Equate (4) and (5):
λ 2 − 9 = − 7 − λ 2 \lambda^2-9 = -7-\lambda^2 λ 2 − 9 = − 7 − λ 2 2 λ 2 = 2 2\lambda^2 = 2 2 λ 2 = 2 λ 2 = 1. \lambda^2 = 1. λ 2 = 1.
Hence
a ⃗ ⋅ b ⃗ = λ 2 − 9 = 1 − 9 = − 8. \vec a\cdot\vec b = \lambda^2-9 = 1-9 = -8. a ⋅ b = λ 2 − 9 = 1 − 9 = − 8.
Therefore,
∣ a ⃗ ⋅ b ⃗ ∣ = 8. \left|\vec a\cdot\vec b\right| = 8. a ⋅ b = 8.
Final answer
8 \boxed{8} 8
The derived answer matches the stored correct answer.