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Question 108 of 162

Q.Find the Vector and Cartesian equations of the plane containing the line x−22=y−23=z−1−2\dfrac{x-2}{2}=\dfrac{y-2}{3}=\dfrac{z-1}{-2} and passing through the point (−1,1,−1)(-1,1,-1).

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2016Subjective· 10mImportance★★★★★
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Take a point and direction vector of the given line, form a second vector to the external point, cross them to get the plane's normal, then write the plane equation.

1. Identify a point and direction vector of the given line x−22=y−23=z−1−2\dfrac{x-2}{2}=\dfrac{y-2}{3}=\dfrac{z-1}{-2}: point A=(2,2,1)A=(2,2,1), direction m⃗=(2,3,−2)\vec m=(2,3,-2).

2. Let B=(−1,1,−1)B=(-1,1,-1) be the given external point that the plane must also pass through.

3. Form the vector AB⃗\vec{AB} (lying in the required plane, along with m⃗\vec m):

AB⃗=B−A=(−1−2, 1−2, −1−1)=(−3,−1,−2)\vec{AB}=B-A=(-1-2,\ 1-2,\ -1-1)=(-3,-1,-2)

4. The plane's normal is n⃗=m⃗×AB⃗\vec n=\vec m\times\vec{AB} (perpendicular to both vectors lying in the plane):

n⃗=∣i⃗j⃗k⃗23−2−3−1−2∣=i⃗(−6−2)−j⃗(−4−6)+k⃗(−2+9)=(−8,10,7)\vec n=\begin{vmatrix}\vec i&\vec j&\vec k\\2&3&-2\\-3&-1&-2\end{vmatrix}=\vec i(-6-2)-\vec j(-4-6)+\vec k(-2+9)=(-8,10,7)

5. Vector equation of the plane through A=(2,2,1)A=(2,2,1) with normal n⃗=(−8,10,7)\vec n=(-8,10,7): …

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