Question:
Let \(\overset{̄}{a},\overset{̄}{b},\overset{̄}{c}\) be the unit vectors such that \(\overset{̄}{a}\) is perpendicular to \(\overset{̄}{b}\) and the angle between \(\overset{̄}{b}\) and \(\overset{̄}{c}\) is \(120^{\circ}\). If \(\overset{̄}{a}+\overset{̄}{c}\) is perpendicular to \(\overset{̄}{b}+\overset{̄}{c}\) then
Let \(\overset{̄}{a},\overset{̄}{b},\overset{̄}{c}\) be the unit vectors such that \(\overset{̄}{a}\) is perpendicular to \(\overset{̄}{b}\) and the angle between \(\overset{̄}{b}\) and \(\overset{̄}{c}\) is \(120^{\circ}\). If \(\overset{̄}{a}+\overset{̄}{c}\) is perpendicular to \(\overset{̄}{b}+\overset{̄}{c}\) then
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Use a perpendicular b, b dot c = -1/2 and (a + c) perpendicular (b + c).
Updated On: Oct 1, 2026
- \((\overset{̄}{a}+\overset{̄}{c})\cdot (\overset{̄}{b}-\overset{̄}{c}) = 1\)
- \((\overset{̄}{a}-\overset{̄}{c})\cdot (\overset{̄}{b}-\overset{̄}{c}) = -2\)
- \((\overset{̄}{a}-\overset{̄}{c})\cdot (\overset{̄}{b}+\overset{̄}{c}) = 1\)
- \((\overset{̄}{a}-\overset{̄}{c})\cdot (\overset{̄}{b}-\overset{̄}{c}) = 2\)
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The Correct Option is D
Solution and Explanation
Step 1: List the dot products:
All three vectors are unit vectors. \(\vec a\cdot\vec b=0\) and \(\vec b\cdot\vec c=\cos120^{\circ}=-\dfrac12\).
Step 2: Use the perpendicular condition:
\((\vec a+\vec c)\cdot(\vec b+\vec c)=0\) gives \(\vec a\cdot\vec b+\vec a\cdot\vec c+\vec c\cdot\vec b+|\vec c|^2=0\). So \(0+\vec a\cdot\vec c-\dfrac12+1=0\), and \(\vec a\cdot\vec c=-\dfrac12\).
Step 3: Test the options:
\((\vec a-\vec c)\cdot(\vec b-\vec c)=\vec a\cdot\vec b-\vec a\cdot\vec c-\vec c\cdot\vec b+|\vec c|^2=0+\dfrac12+\dfrac12+1=2\). So D is true.
Step 4: Why the other options are wrong.
(A): \((\vec a+\vec c)\cdot(\vec b-\vec c)=0+(-\tfrac12)+(-\tfrac12)-1=-2\), not 1. (B): the value is 2, not \(-2\). (C): \((\vec a-\vec c)\cdot(\vec b+\vec c)=\vec a\cdot\vec b+\vec a\cdot\vec c-\vec c\cdot\vec b-1=0-\tfrac12+\tfrac12-1=-1\), not 1.
Final Answer:
The value (a - c) . (b - c) equals 2. \[ \boxed{\text{(D) }(\vec a-\vec c)\cdot(\vec b-\vec c)=2} \]
All three vectors are unit vectors. \(\vec a\cdot\vec b=0\) and \(\vec b\cdot\vec c=\cos120^{\circ}=-\dfrac12\).
Step 2: Use the perpendicular condition:
\((\vec a+\vec c)\cdot(\vec b+\vec c)=0\) gives \(\vec a\cdot\vec b+\vec a\cdot\vec c+\vec c\cdot\vec b+|\vec c|^2=0\). So \(0+\vec a\cdot\vec c-\dfrac12+1=0\), and \(\vec a\cdot\vec c=-\dfrac12\).
Step 3: Test the options:
\((\vec a-\vec c)\cdot(\vec b-\vec c)=\vec a\cdot\vec b-\vec a\cdot\vec c-\vec c\cdot\vec b+|\vec c|^2=0+\dfrac12+\dfrac12+1=2\). So D is true.
Step 4: Why the other options are wrong.
(A): \((\vec a+\vec c)\cdot(\vec b-\vec c)=0+(-\tfrac12)+(-\tfrac12)-1=-2\), not 1. (B): the value is 2, not \(-2\). (C): \((\vec a-\vec c)\cdot(\vec b+\vec c)=\vec a\cdot\vec b+\vec a\cdot\vec c-\vec c\cdot\vec b-1=0-\tfrac12+\tfrac12-1=-1\), not 1.
Final Answer:
The value (a - c) . (b - c) equals 2. \[ \boxed{\text{(D) }(\vec a-\vec c)\cdot(\vec b-\vec c)=2} \]
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