Question:
The equation of the plane passing through the points having position vectors \((\overset{̄}{a}+\overset{̄}{b}),(\overset{̄}{b}+\overset{̄}{c})\) and \((\overset{̄}{a}+\overset{̄}{c})\) is \(\ldots\)
The equation of the plane passing through the points having position vectors \((\overset{̄}{a}+\overset{̄}{b}),(\overset{̄}{b}+\overset{̄}{c})\) and \((\overset{̄}{a}+\overset{̄}{c})\) is \(\ldots\)
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Find a normal using two in-plane vectors, then write r . n = (a+b) . n.
Updated On: Oct 1, 2026
- \(\overset{̄}{r}\cdot (\overset{̄}{a}\times \overset{̄}{b}+\overset{̄}{b}\times \overset{̄}{c}+\overset{̄}{c}\times \overset{̄}{a}) = [\overset{̄}{a} \overset{̄}{b} \overset{̄}{c}]\)
- \(\overset{̄}{r}\cdot (\overset{̄}{a}\times \overset{̄}{b}+\overset{̄}{b}\times \overset{̄}{c}+\overset{̄}{c}\times \overset{̄}{a}) = 2[\overset{̄}{a} \overset{̄}{b} \overset{̄}{c}]\)
- \(\overset{̄}{r}\cdot (\overset{̄}{a}\times \overset{̄}{b}+\overset{̄}{b}\times \overset{̄}{c}+\overset{̄}{a}\times \overset{̄}{c}) = [\overset{̄}{a} \overset{̄}{b} \overset{̄}{c}]\)
- \(\overset{̄}{r}\cdot (\overset{̄}{a}\times \overset{̄}{b}+\overset{̄}{b}\times \overset{̄}{c}+\overset{̄}{a}\times \overset{̄}{c}) = 2[\overset{̄}{a} \overset{̄}{b} \overset{̄}{c}]\)
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Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
A plane through three points has a normal vector equal to the cross product of two vectors lying in it. The equation is \(\bar r\cdot\bar n = \bar p\cdot\bar n\) for any point \(\bar p\) on the plane.
Step 2: Key Formula or Approach:
Let the points be \(P = \bar a + \bar b\), \(Q = \bar b + \bar c\), \(R = \bar a + \bar c\).
Vectors in the plane:
\[ \overrightarrow{PQ} = \bar c - \bar a, \qquad \overrightarrow{PR} = \bar c - \bar b \]
Step 3: Detailed Explanation:
Normal:
\[ \bar n = (\bar c - \bar a)\times(\bar c - \bar b) = \bar c\times\bar c - \bar c\times\bar b - \bar a\times\bar c + \bar a\times\bar b \]
\[ = \bar b\times\bar c + \bar c\times\bar a + \bar a\times\bar b \]
Now evaluate \(\bar p\cdot\bar n\) using \(\bar p = \bar a + \bar b\). Any scalar triple product with a repeated vector is zero:
\[ (\bar a+\bar b)\cdot\left(\bar a\times\bar b + \bar b\times\bar c + \bar c\times\bar a\right) = \bar a\cdot(\bar b\times\bar c) + \bar b\cdot(\bar c\times\bar a) \]
Both terms equal \([\bar a\,\bar b\,\bar c]\), so the total is \(2[\bar a\,\bar b\,\bar c]\).
The plane is
\[ \bar r\cdot\left(\bar a\times\bar b + \bar b\times\bar c + \bar c\times\bar a\right) = 2[\bar a\,\bar b\,\bar c] \]
Option (A) forgets the factor 2. Options (C) and (D) use \(\bar a\times\bar c\) in place of \(\bar c\times\bar a\), which flips the sign of one term and gives a different normal.
Final Answer:
The plane is option (B). \[ \boxed{\bar r\cdot(\bar a\times\bar b+\bar b\times\bar c+\bar c\times\bar a)=2[\bar a\,\bar b\,\bar c] \text{ (B)}} \]
A plane through three points has a normal vector equal to the cross product of two vectors lying in it. The equation is \(\bar r\cdot\bar n = \bar p\cdot\bar n\) for any point \(\bar p\) on the plane.
Step 2: Key Formula or Approach:
Let the points be \(P = \bar a + \bar b\), \(Q = \bar b + \bar c\), \(R = \bar a + \bar c\).
Vectors in the plane:
\[ \overrightarrow{PQ} = \bar c - \bar a, \qquad \overrightarrow{PR} = \bar c - \bar b \]
Step 3: Detailed Explanation:
Normal:
\[ \bar n = (\bar c - \bar a)\times(\bar c - \bar b) = \bar c\times\bar c - \bar c\times\bar b - \bar a\times\bar c + \bar a\times\bar b \]
\[ = \bar b\times\bar c + \bar c\times\bar a + \bar a\times\bar b \]
Now evaluate \(\bar p\cdot\bar n\) using \(\bar p = \bar a + \bar b\). Any scalar triple product with a repeated vector is zero:
\[ (\bar a+\bar b)\cdot\left(\bar a\times\bar b + \bar b\times\bar c + \bar c\times\bar a\right) = \bar a\cdot(\bar b\times\bar c) + \bar b\cdot(\bar c\times\bar a) \]
Both terms equal \([\bar a\,\bar b\,\bar c]\), so the total is \(2[\bar a\,\bar b\,\bar c]\).
The plane is
\[ \bar r\cdot\left(\bar a\times\bar b + \bar b\times\bar c + \bar c\times\bar a\right) = 2[\bar a\,\bar b\,\bar c] \]
Option (A) forgets the factor 2. Options (C) and (D) use \(\bar a\times\bar c\) in place of \(\bar c\times\bar a\), which flips the sign of one term and gives a different normal.
Final Answer:
The plane is option (B). \[ \boxed{\bar r\cdot(\bar a\times\bar b+\bar b\times\bar c+\bar c\times\bar a)=2[\bar a\,\bar b\,\bar c] \text{ (B)}} \]
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