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Vector Algebra question

2003 · Shift 0 · Q94
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Vector Algebra question

2003 · Shift 0 · Q94

JEE MainMathematicsVector AlgebraMCQ+4 / −1
a→ ,b→ ,c→\overrightarrow a \,,\overrightarrow b \,,\overrightarrow ca,b,c are 333 vectors, such that a→+b→+c→=0\overrightarrow a + \overrightarrow b + \overrightarrow c = 0a+b+c=0, ∣a→∣=1   ∣b→∣=2,   ∣c→∣=3,\left| {\overrightarrow a } \right| = 1\,\,\,\left| {\overrightarrow b } \right| = 2,\,\,\,\left| {\overrightarrow c } \right| = 3,​a​=1​b​=2,​c​=3,, then a→.b→+b→.c→+c→.a→{\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a }a.b+b.c+c.a is equal to :
  1. A
    111
  2. B
    000
  3. C
    −7-7−7
  4. D
    777
View written solutionFree

Correct answer: C

  1. We are given a⃗+b⃗+c⃗=0\vec a + \vec b + \vec c = 0a+b+c=0 with magnitudes ∣a⃗∣=1,∣b⃗∣=2,∣c⃗∣=3.|\vec a|=1,\quad |\vec b|=2,\quad |\vec c|=3.∣a∣=1,∣b∣=2,∣c∣=3.

  2. To find a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗,\vec a\cdot \vec b + \vec b\cdot \vec c + \vec c\cdot \vec a,a⋅b+b⋅c+c⋅a, use the identity: (a⃗+b⃗+c⃗)2=∣a⃗∣2+∣b⃗∣2+∣c⃗∣2+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗).(\vec a+\vec b+\vec c)^2 = |\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).(a+b+c)2=∣a∣2+∣b∣2+∣c∣2+2(a⋅b+b⋅c+c⋅a).

  3. Since a⃗+b⃗+c⃗=0,\vec a+\vec b+\vec c=0,a+b+c=0, its square is also zero: 0=∣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).

  4. Substitute the given magnitudes: 0=12+22+32+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗).0 = 1^2 + 2^2 + 3^2 + 2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a).0=12+22+32+2(a⋅b+b⋅c+c⋅a). 0=1+4+9+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗).0 = 1 + 4 + 9 + 2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a).0=1+4+9+2(a⋅b+b⋅c+c⋅a). 0=14+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗).0 = 14 + 2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a).0=14+2(a⋅b+b⋅c+c⋅a).

  5. Therefore, 2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=−14,2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a) = -14,2(a⋅b+b⋅c+c⋅a)=−14, so a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗=−7.\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a = -7.a⋅b+b⋅c+c⋅a=−7.

  6. Hence the correct option is: C: −7\boxed{\text{C: }-7}C: −7​

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