Exams Joint Entrance Examination Mathematics Vector Algebra Question 10 Question 10 of 379 Medium MCQ
For three unit vectors a ⃗ , b ⃗ , c ⃗ \vec{a},\vec{b},\vec{c} a , b , c satisfying
∣ a ⃗ − b ⃗ ∣ 2 + ∣ b ⃗ − c ⃗ ∣ 2 + ∣ c ⃗ − a ⃗ ∣ 2 = 9 ∣ 2 a ⃗ + k b ⃗ + k c ⃗ ∣ = 3 |\vec{a}-\vec{b}{|}^{2}+|\vec{b}-\vec{c}{|}^{2}+|\vec{c}-\vec{a}{|}^{2}=9|2\vec{a}+k\vec{b}+k\vec{c}|=3 ∣ a − b ∣ 2 + ∣ b − c ∣ 2 + ∣ c − a ∣ 2 = 9∣2 a + k b + k c ∣ = 3
the positive value of k is :
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Given, ∣ a ⃗ ∣ = ∣ b ⃗ ∣ = ∣ c ⃗ ∣ = 1 |\vec{a}|=|\vec{b}|=|\vec{c}|=1 ∣ a ∣ = ∣ b ∣ = ∣ c ∣ = 1
∣ a ⃗ − b ⃗ ∣ 2 + ∣ b ⃗ − c ⃗ ∣ 2 + ∣ c ⃗ − a ⃗ ∣ 2 = 9 |\vec{a}-\vec{b}{|}^{2}+|\vec{b}-\vec{c}{|}^{2}+|\vec{c}-\vec{a}{|}^{2}=9 ∣ a − b ∣ 2 + ∣ b − c ∣ 2 + ∣ c − a ∣ 2 = 9
⇒ \Rightarrow ⇒ ∣ a ⃗ ∣ 2 + ∣ b ⃗ ∣ 2 − 2 a ⃗ ⋅ b ⃗ + ∣ b ⃗ ∣ 2 + ∣ c ⃗ ∣ 2 − 2 b ⃗ ⋅ c ⃗ + ∣ c ⃗ ∣ 2 + ∣ a ⃗ ∣ 2 − 2 c ⃗ ⋅ a ⃗ = 9 |\vec{a}{|}^{2}+|\vec{b}{|}^{2}-2\vec{a}\cdot \vec{b}+|\vec{b}{|}^{2}+|\vec{c}{|}^{2}-2\vec{b}\cdot \vec{c}+|\vec{c}{|}^{2}+|\vec{a}{|}^{2}-2\vec{c}\cdot \vec{a}=9 ∣ a ∣ 2 + ∣ b ∣ 2 − 2 a ⋅ b + ∣ b ∣ 2 + ∣ c ∣ 2 − 2 b ⋅ c + ∣ c ∣ 2 + ∣ a ∣ 2 − 2 c ⋅ a = 9
⇒ \Rightarrow ⇒ 6 + 2 ( a ⃗ ⋅ b ⃗ + b ⃗ ⋅ c ⃗ + c ⃗ ⋅ a ⃗ ) = 9 ∣ a ⃗ ∣ = ∣ b ⃗ ∣ = ∣ c ⃗ ∣ = 1 6+2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})=9\, {|\vec{a}|=|\vec{b}|=|\vec{c}|=1} 6 + 2 ( a ⋅ b + b ⋅ c + c ⋅ a ) = 9 ∣ a ∣ = ∣ b ∣ = ∣ c ∣ = 1
a ⃗ ⋅ b ⃗ + b ⃗ ⋅ c ⃗ + c ⃗ ⋅ a ⃗ = − 3 2 \vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}=-\frac{3}{2} a ⋅ b + b ⋅ c + c ⋅ a = − 2 3
also,
∣ a ⃗ + b ⃗ + c ⃗ ∣ 2 = 3 + 2 ( a ⃗ ⋅ b ⃗ + b ⃗ ⋅ c ⃗ + c ⃗ ⋅ a ⃗ ) |\vec{a}+\vec{b}+\vec{c}{|}^{2}=3+2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}) ∣ a + b + c ∣ 2 = 3 + 2 ( a ⋅ b + b ⋅ c + c ⋅ a )
= 3 + 2 ( − 3 2 ) = 3 − 3 = 0 =3+2(-\frac{3}{2})=3-3=0 = 3 + 2 ( − 2 3 ) = 3 − 3 = 0
So
a ⃗ + b ⃗ + c ⃗ = 0 \vec{a}+\vec{b}+\vec{c}=0 a + b + c = 0
⇒ \Rightarrow ⇒ a ⃗ + b ⃗ = − c ⃗ \vec{a}+\vec{b}=-\vec{c} a + b = − c
⇒ \Rightarrow ⇒ ∣ a ⃗ + b ⃗ ∣ 2 = ∣ c ⃗ ∣ 2 = 1 |\vec{a}+\vec{b}{|}^{2}=|\vec{c}{|}^{2}=1 ∣ a + b ∣ 2 = ∣ c ∣ 2 = 1
⇒ \Rightarrow ⇒ ∣ a ⃗ ∣ 2 + ∣ b ⃗ ∣ 2 + 2 a ⃗ ⋅ b ⃗ = 1 |\vec{a}{|}^{2}+|\vec{b}{|}^{2}+2\vec{a}\cdot \vec{b}=1 ∣ a ∣ 2 + ∣ b ∣ 2 + 2 a ⋅ b = 1
⇒ \Rightarrow ⇒ 1 + 1 + 2 a ⃗ ⋅ b ⃗ = 1 1+1+2\vec{a}\cdot \vec{b}=1 1 + 1 + 2 a ⋅ b = 1
⇒ \Rightarrow ⇒ a ⃗ ⋅ b ⃗ = − 1 2 \vec{a}\cdot \vec{b}=-\frac{1}{2} a ⋅ b = − 2 1
Similarly,
b ⃗ ⋅ c ⃗ = − 1 2 a ⃗ ⋅ c ⃗ = − 1 2 \vec{b}\cdot \vec{c}=-\frac{1}{2}\, \, \vec{a}\cdot \vec{c}=-\frac{1}{2} b ⋅ c = − 2 1 a ⋅ c = − 2 1
so,
a ⃗ ⋅ b ⃗ = b ⃗ ⋅ c ⃗ = c ⃗ ⋅ a ⃗ = − 1 2 \vec{a}\cdot \vec{b}=\vec{b}\cdot \vec{c}=\vec{c}\cdot \vec{a}=-\frac{1}{2} a ⋅ b = b ⋅ c = c ⋅ a = − 2 1
It is given that
∣ 2 a ⃗ + k b ⃗ + k c ⃗ ∣ = 3 |2\vec{a}+k\vec{b}+k\vec{c}|=3 ∣2 a + k b + k c ∣ = 3
⇒ \Rightarrow ⇒ ∣ 2 a ⃗ + k b ⃗ + k c ⃗ ∣ 2 = 9 |2\vec{a}+k\vec{b}+k\vec{c}{|}^{2}=9 ∣2 a + k b + k c ∣ 2 = 9
⇒ \Rightarrow ⇒ 4 ∣ a ⃗ ∣ 2 + k 2 ∣ b ⃗ ∣ 2 + k 2 ∣ c ⃗ ∣ 2 + 4 k a ⃗ ⋅ b ⃗ + 2 k 2 b ⃗ ⋅ c ⃗ + 4 k a ⃗ ⋅ c ⃗ = 9 4|\vec{a}{|}^{2}+{k}^{2}|\vec{b}{|}^{2}+{k}^{2}|\vec{c}{|}^{2}+4k\vec{a}\cdot \vec{b}+2{k}^{2}\vec{b}\cdot \vec{c}+4k\vec{a}\cdot \vec{c}=9 4∣ a ∣ 2 + k 2 ∣ b ∣ 2 + k 2 ∣ c ∣ 2 + 4 k a ⋅ b + 2 k 2 b ⋅ c + 4 k a ⋅ c = 9
⇒ \Rightarrow ⇒ 4 ( 1 ) 2 + k 2 ( 1 ) 2 + k 2 ( 1 ) 2 + 4 k ( − 1 2 ) + 2 k 2 ( − 1 2 ) + 4 k ( − 1 2 ) = 9 4(1{)}^{2}+{k}^{2}(1{)}^{2}+{k}^{2}(1{)}^{2}+4k(-\frac{1}{2})+2{k}^{2}(-\frac{1}{2})+4k(-\frac{1}{2})=9 4 ( 1 ) 2 + k 2 ( 1 ) 2 + k 2 ( 1 ) 2 + 4 k ( − 2 1 ) + 2 k 2 ( − 2 1 ) + 4 k ( − 2 1 ) = 9
⇒ \Rightarrow ⇒ 4 + 2 k 2 − 2 k − k 2 − 2 k = 9 4+2{k}^{2}-2k-{k}^{2}-2k=9 4 + 2 k 2 − 2 k − k 2 − 2 k = 9
⇒ \Rightarrow ⇒ k 2 − 4 k − 5 = 0 {k}^{2}-4k-5=0 k 2 − 4 k − 5 = 0
⇒ \Rightarrow ⇒ ( k − 5 ) ( k + 1 ) = 0 (k-5)(k+1)=0 ( k − 5 ) ( k + 1 ) = 0
⇒ \Rightarrow ⇒ k = 5 , − 1 k=5,-1 k = 5 , − 1 Answer the question to unlock