Write the position vectors
Given
O A ⃗ = i ^ + j ^ + k ^ = ( 1 , 1 , 1 ) , O B ⃗ = 2 i ^ + j ^ + 3 k ^ = ( 2 , 1 , 3 ) \vec{OA}=\hat i+\hat j+\hat k=(1,1,1),
\qquad
\vec{OB}=2\hat i+\hat j+3\hat k=(2,1,3) O A = i ^ + j ^ + k ^ = ( 1 , 1 , 1 ) , O B = 2 i ^ + j ^ + 3 k ^ = ( 2 , 1 , 3 ) Point P P P A B AB A B λ : 1 \lambda:1 λ : 1
A P : P B = λ : 1. AP:PB=\lambda:1. A P : P B = λ : 1. Hence, by section formula,
O P ⃗ = O A ⃗ + λ O B ⃗ λ + 1 . \vec{OP}=\frac{\vec{OA}+\lambda\vec{OB}}{\lambda+1}. O P = λ + 1 O A + λ O B . Substituting O A ⃗ \vec{OA} O A O B ⃗ \vec{OB} O B
O P ⃗ = ( 1 , 1 , 1 ) + λ ( 2 , 1 , 3 ) λ + 1 = ( 1 + 2 λ λ + 1 , 1 , 1 + 3 λ λ + 1 ) . \vec{OP}=\frac{(1,1,1)+\lambda(2,1,3)}{\lambda+1}
=\left(\frac{1+2\lambda}{\lambda+1},\,1,\,\frac{1+3\lambda}{\lambda+1}\right). O P = λ + 1 ( 1 , 1 , 1 ) + λ ( 2 , 1 , 3 ) = ( λ + 1 1 + 2 λ , 1 , λ + 1 1 + 3 λ ) . Let
O P ⃗ = ( x , 1 , z ) , x = 1 + 2 λ λ + 1 , z = 1 + 3 λ λ + 1 . \vec{OP}=(x,1,z), \quad x=\frac{1+2\lambda}{\lambda+1}, \quad z=\frac{1+3\lambda}{\lambda+1}. O P = ( x , 1 , z ) , x = λ + 1 1 + 2 λ , z = λ + 1 1 + 3 λ .
Compute O B ⃗ ⋅ O P ⃗ \vec{OB}\cdot\vec{OP} O B ⋅ O P
O B ⃗ ⋅ O P ⃗ = ( 2 , 1 , 3 ) ⋅ ( x , 1 , z ) = 2 x + 1 + 3 z . \vec{OB}\cdot\vec{OP}=(2,1,3)\cdot(x,1,z)=2x+1+3z. O B ⋅ O P = ( 2 , 1 , 3 ) ⋅ ( x , 1 , z ) = 2 x + 1 + 3 z . Now,
2 x = 2 + 4 λ λ + 1 , 3 z = 3 + 9 λ λ + 1 . 2x=\frac{2+4\lambda}{\lambda+1},
\qquad
3z=\frac{3+9\lambda}{\lambda+1}. 2 x = λ + 1 2 + 4 λ , 3 z = λ + 1 3 + 9 λ . So,
O B ⃗ ⋅ O P ⃗ = 1 + 5 + 13 λ λ + 1 = λ + 1 + 5 + 13 λ λ + 1 = 14 λ + 6 λ + 1 . \vec{OB}\cdot\vec{OP}
=1+\frac{5+13\lambda}{\lambda+1}
=\frac{\lambda+1+5+13\lambda}{\lambda+1}
=\frac{14\lambda+6}{\lambda+1}. O B ⋅ O P = 1 + λ + 1 5 + 13 λ = λ + 1 λ + 1 + 5 + 13 λ = λ + 1 14 λ + 6 .
Compute ∣ O A ⃗ × O P ⃗ ∣ 2 |\vec{OA}\times\vec{OP}|^2 ∣ O A × O P ∣ 2
We use
O A ⃗ = ( 1 , 1 , 1 ) , O P ⃗ = ( x , 1 , z ) . \vec{OA}=(1,1,1), \quad \vec{OP}=(x,1,z). O A = ( 1 , 1 , 1 ) , O P = ( x , 1 , z ) . Then
O A ⃗ × O P ⃗ = ∣ i ^ j ^ k ^ 1 1 1 x 1 z ∣ = ( z − 1 ) i ^ − ( z − x ) j ^ + ( 1 − x ) k ^ . \vec{OA}\times\vec{OP}
=\begin{vmatrix}
\hat i & \hat j & \hat k\\
1 & 1 & 1\\
x & 1 & z
\end{vmatrix}
=(z-1)\hat i-(z-x)\hat j+(1-x)\hat k. O A × O P = i ^ 1 x j ^ 1 1 k ^ 1 z = ( z − 1 ) i ^ − ( z − x ) j ^ + ( 1 − x ) k ^ . Thus,
∣ O A ⃗ × O P ⃗ ∣ 2 = ( z − 1 ) 2 + ( z − x ) 2 + ( 1 − x ) 2 . |\vec{OA}\times\vec{OP}|^2=(z-1)^2+(z-x)^2+(1-x)^2. ∣ O A × O P ∣ 2 = ( z − 1 ) 2 + ( z − x ) 2 + ( 1 − x ) 2 . Now,
z − 1 = 1 + 3 λ λ + 1 − 1 = 2 λ λ + 1 , z-1=\frac{1+3\lambda}{\lambda+1}-1=\frac{2\lambda}{\lambda+1}, z − 1 = λ + 1 1 + 3 λ − 1 = λ + 1 2 λ , 1 − x = 1 − 1 + 2 λ λ + 1 = − λ λ + 1 , 1-x=1-\frac{1+2\lambda}{\lambda+1}=\frac{-\lambda}{\lambda+1}, 1 − x = 1 − λ + 1 1 + 2 λ = λ + 1 − λ , z − x = 1 + 3 λ − ( 1 + 2 λ ) λ + 1 = λ λ + 1 . z-x=\frac{1+3\lambda-(1+2\lambda)}{\lambda+1}=\frac{\lambda}{\lambda+1}. z − x = λ + 1 1 + 3 λ − ( 1 + 2 λ ) = λ + 1 λ . Therefore,
∣ O A ⃗ × O P ⃗ ∣ 2 = ( 2 λ λ + 1 ) 2 + ( λ λ + 1 ) 2 + ( − λ λ + 1 ) 2 = 4 λ 2 + λ 2 + λ 2 ( λ + 1 ) 2 = 6 λ 2 ( λ + 1 ) 2 . |\vec{OA}\times\vec{OP}|^2
=\left(\frac{2\lambda}{\lambda+1}\right)^2+\left(\frac{\lambda}{\lambda+1}\right)^2+\left(\frac{-\lambda}{\lambda+1}\right)^2
=\frac{4\lambda^2+\lambda^2+\lambda^2}{(\lambda+1)^2}
=\frac{6\lambda^2}{(\lambda+1)^2}. ∣ O A × O P ∣ 2 = ( λ + 1 2 λ ) 2 + ( λ + 1 λ ) 2 + ( λ + 1 − λ ) 2 = ( λ + 1 ) 2 4 λ 2 + λ 2 + λ 2 = ( λ + 1 ) 2 6 λ 2 . So,
3 ∣ O A ⃗ × O P ⃗ ∣ 2 = 18 λ 2 ( λ + 1 ) 2 . 3|\vec{OA}\times\vec{OP}|^2=\frac{18\lambda^2}{(\lambda+1)^2}. 3∣ O A × O P ∣ 2 = ( λ + 1 ) 2 18 λ 2 .
Use the given condition
Given
O B ⃗ ⋅ O P ⃗ − 3 ∣ O A ⃗ × O P ⃗ ∣ 2 = 6. \vec{OB}\cdot\vec{OP}-3|\vec{OA}\times\vec{OP}|^2=6. O B ⋅ O P − 3∣ O A × O P ∣ 2 = 6. Substitute the expressions:
14 λ + 6 λ + 1 − 18 λ 2 ( λ + 1 ) 2 = 6. \frac{14\lambda+6}{\lambda+1}-\frac{18\lambda^2}{(\lambda+1)^2}=6. λ + 1 14 λ + 6 − ( λ + 1 ) 2 18 λ 2 = 6. Multiply by ( λ + 1 ) 2 (\lambda+1)^2 ( λ + 1 ) 2
( 14 λ + 6 ) ( λ + 1 ) − 18 λ 2 = 6 ( λ + 1 ) 2 . (14\lambda+6)(\lambda+1)-18\lambda^2=6(\lambda+1)^2. ( 14 λ + 6 ) ( λ + 1 ) − 18 λ 2 = 6 ( λ + 1 ) 2 . Expand:
14 λ 2 + 20 λ + 6 − 18 λ 2 = 6 λ 2 + 12 λ + 6. 14\lambda^2+20\lambda+6-18\lambda^2=6\lambda^2+12\lambda+6. 14 λ 2 + 20 λ + 6 − 18 λ 2 = 6 λ 2 + 12 λ + 6. − 4 λ 2 + 20 λ + 6 = 6 λ 2 + 12 λ + 6. -4\lambda^2+20\lambda+6=6\lambda^2+12\lambda+6. − 4 λ 2 + 20 λ + 6 = 6 λ 2 + 12 λ + 6. − 10 λ 2 + 8 λ = 0. -10\lambda^2+8\lambda=0. − 10 λ 2 + 8 λ = 0. 2 λ ( − 5 λ + 4 ) = 0. 2\lambda(-5\lambda+4)=0. 2 λ ( − 5 λ + 4 ) = 0. Since λ > 0 \lambda>0 λ > 0
− 5 λ + 4 = 0 ⇒ λ = 4 5 = 0.8. -5\lambda+4=0
\quad\Rightarrow\quad
\lambda=\frac45=0.8. − 5 λ + 4 = 0 ⇒ λ = 5 4 = 0.8.
Final answer
λ = 0.8 \boxed{\lambda=0.8} λ = 0.8