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Vector Algebra question

2002 · Shift 0 · Q91
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  5. /2002 · Shift 0 · Q91

Vector Algebra question

2002 · Shift 0 · Q91

JEE MainMathematicsVector AlgebraMCQ+4 / −1
If ∣a→∣=5,∣b→∣=4,∣c→∣=3\left| {\overrightarrow a } \right| = 5,\left| {\overrightarrow b } \right| = 4,\left| {\overrightarrow c } \right| = 3​a​=5,​b​=4,​c​=3 thus what will be the value of ∣a→.b→+b→.c→+c→.a→∣,\left| {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right|,​a.b+b.c+c.a​, given that a→+b→+c→=0\overrightarrow a + \overrightarrow b + \overrightarrow c = 0a+b+c=0 :
  1. A
    252525
  2. B
    505050
  3. C
    −25-25−25
  4. D
    −50-50−50
View written solutionFree

Correct answer: A

  1. We are given ∣a⃗∣=5,∣b⃗∣=4,∣c⃗∣=3,|\vec a|=5,\quad |\vec b|=4,\quad |\vec c|=3,∣a∣=5,∣b∣=4,∣c∣=3, and a⃗+b⃗+c⃗=0.\vec a+\vec b+\vec c=0.a+b+c=0.

  2. Use the identity ∣a⃗+b⃗+c⃗∣2=(a⃗+b⃗+c⃗)⋅(a⃗+b⃗+c⃗).|\vec a+\vec b+\vec c|^2=(\vec a+\vec b+\vec c)\cdot(\vec a+\vec b+\vec c).∣a+b+c∣2=(a+b+c)⋅(a+b+c). Since a⃗+b⃗+c⃗=0\vec a+\vec b+\vec c=0a+b+c=0, we have ∣a⃗+b⃗+c⃗∣2=0.|\vec a+\vec b+\vec c|^2=0.∣a+b+c∣2=0.

  3. Expand the dot product: ∣a⃗∣2+∣b⃗∣2+∣c⃗∣2+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=0.|\vec a|^2+|\vec b|^2+|\vec c|^2+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0.∣a∣2+∣b∣2+∣c∣2+2(a⋅b+b⋅c+c⋅a)=0.

  4. Substitute the magnitudes: 52+42+32+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=0.5^2+4^2+3^2+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0.52+42+32+2(a⋅b+b⋅c+c⋅a)=0. 25+16+9+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=0.25+16+9+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0.25+16+9+2(a⋅b+b⋅c+c⋅a)=0. 50+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=0.50+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0.50+2(a⋅b+b⋅c+c⋅a)=0.

  5. Therefore, 2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=−50,2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=-50,2(a⋅b+b⋅c+c⋅a)=−50, so a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗=−25.\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a=-25.a⋅b+b⋅c+c⋅a=−25.

  6. The question asks for ∣a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗∣.\left|\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a\right|.​a⋅b+b⋅c+c⋅a​. Hence, ∣−25∣=25.\left|-25\right|=25.∣−25∣=25.

  7. So the correct option is A:25.\boxed{A: 25}.A:25​.

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