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Vector Algebra question

2014 · Shift 1 · Q30
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  5. /2014 · Shift 1 · Q30

Vector Algebra question

2014 · Shift 1 · Q30

JEE AdvancedMathematicsVector AlgebraMultiple correct+3 / −1
Let x→,y→\overrightarrow x ,\overrightarrow yx,y​ and z→\overrightarrow zz be three vectors each of magnitude 2\sqrt 22​ and the angle between each pair of them is π3{\pi \over 3}3π​. If a→\overrightarrow aa is a non-zero vector perpendicular to x→\overrightarrow xx and y→×z→\overrightarrow y \times \overrightarrow zy​×z and b→\overrightarrow bb is a non-zero vector perpendicular to y→\overrightarrow yy​ and z→×x→,\overrightarrow z \times \overrightarrow x ,z×x, then
  1. A
    b→=(b→ . z→)(z→−x→)\overrightarrow b = \left( {\overrightarrow b \,.\,\overrightarrow z } \right)\left( {\overrightarrow z - \overrightarrow x } \right)b=(b.z)(z−x)
  2. B
    a→=(a→ . y→)(y→−z→)\overrightarrow a = \left( {\overrightarrow a \,.\,\overrightarrow y } \right)\left( {\overrightarrow y - \overrightarrow z } \right)a=(a.y​)(y​−z)
  3. C
    a→ . b→=−(a→ . y→)(b→ . z→)\overrightarrow a \,.\,\overrightarrow b = - \left( {\overrightarrow a \,.\,\overrightarrow y } \right)\left( {\overrightarrow b \,.\,\overrightarrow z } \right)a.b=−(a.y​)(b.z)
  4. D
    a→=(a→ . y→)(z→−y→)\overrightarrow a = \left( {\overrightarrow a \,.\,\overrightarrow y } \right)\left( {\overrightarrow z - \overrightarrow y } \right)a=(a.y​)(z−y​)
View written solutionFree

Correct answer: A, B, C

  1. Given data

Each vector has magnitude 2\sqrt{2}2​ and angle π/3\pi/3π/3 between each pair. So

\vec x|=|\vec y|=|\vec z|=\sqrt2, \qquad \vec x\cdot \vec y=\vec y\cdot \vec z=\vec z\cdot \vec x =|\vec x||\vec y|\cos\frac\pi3=2\cdot \frac12=1.$$ Thus, $$\vec x\cdot \vec y=\vec y\cdot \vec z=\vec z\cdot \vec x=1.$$ --- 2. **Find the direction of $\vec a$** We are given that $\vec a$ is perpendicular to $\vec x$ and to $\vec y\times \vec z$. A vector perpendicular to $\vec y\times \vec z$ lies in the plane of $\vec y$ and $\vec z$. Hence $\vec a$ lies in the plane spanned by $\vec y,\vec z$. So write $$\vec a=\alpha \vec y+\beta \vec z.$$ Also $\vec a\perp \vec x$, so $$\vec a\cdot \vec x=0$$ $$\Rightarrow (\alpha \vec y+\beta \vec z)\cdot \vec x=0$$ $$\Rightarrow \alpha (\vec y\cdot \vec x)+\beta (\vec z\cdot \vec x)=0$$ $$\Rightarrow \alpha+\beta=0.$$ So $\beta=-\alpha$, and therefore $$\vec a=\alpha(\vec y-\vec z).$$ Now compute $\vec a\cdot \vec y$: $$\vec a\cdot \vec y=\alpha(\vec y-\vec z)\cdot \vec y =\alpha(\vec y\cdot \vec y-\vec z\cdot \vec y) =\alpha(2-1)=\alpha.$$ Hence $$\boxed{\vec a=(\vec a\cdot \vec y)(\vec y-\vec z).}$$ So **Option B is correct**. Also, $$(\vec a\cdot \vec y)(\vec z-\vec y)=-(\vec a\cdot \vec y)(\vec y-\vec z)=-\vec a,$$ so this is not equal to $\vec a$ since $\vec a\neq 0$. Thus **Option D is false**. --- 3. **Find the direction of $\vec b$** $\vec b$ is perpendicular to $\vec y$ and to $\vec z\times \vec x$. A vector perpendicular to $\vec z\times \vec x$ lies in the plane of $\vec z$ and $\vec x$. So write $$\vec b=\lambda \vec z+\mu \vec x.$$ Since $\vec b\perp \vec y$, $$\vec b\cdot \vec y=0$$ $$\Rightarrow (\lambda \vec z+\mu \vec x)\cdot \vec y=0$$ $$\Rightarrow \lambda(\vec z\cdot \vec y)+\mu(\vec x\cdot \vec y)=0$$ $$\Rightarrow \lambda+\mu=0.$$ So $\mu=-\lambda$, hence $$\vec b=\lambda(\vec z-\vec x).$$ Now $$\vec b\cdot \vec z=\lambda(\vec z-\vec x)\cdot \vec z =\lambda(\vec z\cdot \vec z-\vec x\cdot \vec z) =\lambda(2-1)=\lambda.$$ Therefore $$\boxed{\vec b=(\vec b\cdot \vec z)(\vec z-\vec x).}$$ So **Option A is correct**. --- 4. **Check Option C** From above, $$\vec a=(\vec a\cdot \vec y)(\vec y-\vec z), \qquad \vec b=(\vec b\cdot \vec z)(\vec z-\vec x).$$ Therefore, $$\vec a\cdot \vec b=(\vec a\cdot \vec y)(\vec b\cdot \vec z)\, (\vec y-\vec z)\cdot(\vec z-\vec x).$$ Now evaluate $$ (\vec y-\vec z)\cdot(\vec z-\vec x) =\vec y\cdot \vec z-\vec y\cdot \vec x-\vec z\cdot \vec z+\vec z\cdot \vec x.$$ Substitute values: $$=1-1-2+1=-1.$$ So $$\vec a\cdot \vec b=-(\vec a\cdot \vec y)(\vec b\cdot \vec z).$$ Hence **Option C is correct**. --- 5. **Final evaluation of options** - **A:** Correct - **B:** Correct - **C:** Correct - **D:** False So the correct set is $$\boxed{A,\ B,\ C}. $$ --- 6. **Comparison with stored answer** Stored correct answer: **A, B, C** Our derived answer matches exactly.
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