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Vector Algebra question

2011 · Shift 2 · Q31
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  5. /2011 · Shift 2 · Q31

Vector Algebra question

2011 · Shift 2 · Q31

JEE AdvancedMathematicsVector AlgebraNumerical+4 / −1
Let a→=−i^−k^,b→=−i^+j^\overrightarrow a = - \widehat i - \widehat k,\overrightarrow b = - \widehat i + \widehat ja=−i−k,b=−i+j​ and c→=i^+2j^+3k^\overrightarrow c = \widehat i + 2\widehat j + 3\widehat kc=i+2j​+3k be three given vectors. If r→\overrightarrow rr is a vector such that r→×b→=c→×b→\overrightarrow r \times \overrightarrow b = \overrightarrow c \times \overrightarrow br×b=c×b and r→.a→=0,\overrightarrow r .\overrightarrow a = 0,r.a=0, then the value of r→.b→\overrightarrow r .\overrightarrow br.b is
Numerical answer
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Correct answer: 9

Step-by-step Solution:

  1. Analyze the given vector equation: We are given the equation r→×b→=c→×b→\overrightarrow r \times \overrightarrow b = \overrightarrow c \times \overrightarrow br×b=c×b. This can be rearranged as: r→×b→−c→×b→=0→\overrightarrow r \times \overrightarrow b - \overrightarrow c \times \overrightarrow b = \overrightarrow 0r×b−c×b=0 (r→−c→)×b→=0→(\overrightarrow r - \overrightarrow c) \times \overrightarrow b = \overrightarrow 0(r−c)×b=0 This equation implies that the vector (r→−c→)(\overrightarrow r - \overrightarrow c)(r−c) is parallel to the vector b→\overrightarrow bb. Two vectors are parallel if one is a scalar multiple of the other. Therefore, we can write: r→−c→=λb→\overrightarrow r - \overrightarrow c = \lambda \overrightarrow br−c=λb where λ\lambdaλ is a scalar. From this, we get an expression for r→\overrightarrow rr: r→=c→+λb→⋯(1)\overrightarrow r = \overrightarrow c + \lambda \overrightarrow b \quad \cdots(1)r=c+λb⋯(1)

  2. Use the second given condition: We are also given that r→.a→=0\overrightarrow r . \overrightarrow a = 0r.a=0. Substitute the expression for r→\overrightarrow rr from equation (1) into this condition: (c→+λb→).a→=0(\overrightarrow c + \lambda \overrightarrow b) . \overrightarrow a = 0(c+λb).a=0 c→.a→+λ(b→.a→)=0⋯(2)\overrightarrow c . \overrightarrow a + \lambda (\overrightarrow b . \overrightarrow a) = 0 \quad \cdots(2)c.a+λ(b.a)=0⋯(2)

  3. Calculate the required dot products: The given vectors are: a→=−i^−k^=⟨−1,0,−1⟩\overrightarrow a = - \widehat i - \widehat k = \langle -1, 0, -1 \ranglea=−i−k=⟨−1,0,−1⟩ b→=−i^+j^=⟨−1,1,0⟩\overrightarrow b = - \widehat i + \widehat j = \langle -1, 1, 0 \rangleb=−i+j​=⟨−1,1,0⟩ c→=i^+2j^+3k^=⟨1,2,3⟩\overrightarrow c = \widehat i + 2\widehat j + 3\widehat k = \langle 1, 2, 3 \ranglec=i+2j​+3k=⟨1,2,3⟩

    Now, we calculate c→.a→\overrightarrow c . \overrightarrow ac.a and b→.a→\overrightarrow b . \overrightarrow ab.a: c→.a→=(1)(−1)+(2)(0)+(3)(−1)=−1−3=−4\overrightarrow c . \overrightarrow a = (1)(-1) + (2)(0) + (3)(-1) = -1 - 3 = -4c.a=(1)(−1)+(2)(0)+(3)(−1)=−1−3=−4 b→.a→=(−1)(−1)+(1)(0)+(0)(−1)=1+0=1\overrightarrow b . \overrightarrow a = (-1)(-1) + (1)(0) + (0)(-1) = 1 + 0 = 1b.a=(−1)(−1)+(1)(0)+(0)(−1)=1+0=1

  4. Solve for the scalar λ\lambdaλ: Substitute the values of the dot products back into equation (2): −4+λ(1)=0-4 + \lambda (1) = 0−4+λ(1)=0 λ=4\lambda = 4λ=4

  5. Calculate the final value r→.b→\overrightarrow r . \overrightarrow br.b: The question asks for the value of r→.b→\overrightarrow r . \overrightarrow br.b. Using the expression for r→\overrightarrow rr from equation (1), we get: r→.b→=(c→+λb→).b→\overrightarrow r . \overrightarrow b = (\overrightarrow c + \lambda \overrightarrow b) . \overrightarrow br.b=(c+λb).b r→.b→=c→.b→+λ(b→.b→)\overrightarrow r . \overrightarrow b = \overrightarrow c . \overrightarrow b + \lambda (\overrightarrow b . \overrightarrow b)r.b=c.b+λ(b.b) We need to calculate c→.b→\overrightarrow c . \overrightarrow bc.b and b→.b→\overrightarrow b . \overrightarrow bb.b: c→.b→=(1)(−1)+(2)(1)+(3)(0)=−1+2=1\overrightarrow c . \overrightarrow b = (1)(-1) + (2)(1) + (3)(0) = -1 + 2 = 1c.b=(1)(−1)+(2)(1)+(3)(0)=−1+2=1 b→.b→=∣b→∣2=(−1)2+(1)2+(0)2=1+1=2\overrightarrow b . \overrightarrow b = |\overrightarrow b|^2 = (-1)^2 + (1)^2 + (0)^2 = 1 + 1 = 2b.b=∣b∣2=(−1)2+(1)2+(0)2=1+1=2 Now substitute the values of c→.b→\overrightarrow c . \overrightarrow bc.b, b→.b→\overrightarrow b . \overrightarrow bb.b, and λ=4\lambda = 4λ=4: r→.b→=1+(4)(2)=1+8=9\overrightarrow r . \overrightarrow b = 1 + (4)(2) = 1 + 8 = 9r.b=1+(4)(2)=1+8=9

Thus, the value of r→.b→\overrightarrow r . \overrightarrow br.b is 9.

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