Let vec{a} = 2hat{i}-hat{j}-hat{k},vec{b} = 5 | vedclass

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(B) Given position vectors are $\vec{a} = 2\hat{i}-\hat{j}-\hat{k}$,$\vec{b} = 5\hat{i}+\hat{j}-2\hat{k}$,and $\vec{c} = -13\hat{i}-11\hat{j}+4\hat{k}

Vedclass · Accepted Answer

$-1$

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(B) Given position vectors are $\vec{a} = 2\hat{i}-\hat{j}-\hat{k}$,$\vec{b} = 5\hat{i}+\hat{j}-2\hat{k}$,and $\vec{c} = -13\hat{i}-11\hat{j}+4\hat{k}$.First,calculate the vectors $\vec{AB}$,$\vec{BC}$,$\vec{AC}$,and $\vec{CB}$:$\vec{AB} = \vec{b} - \vec{a} = (5-2)\hat{i} + (1-(-1))\hat{j} + (-2-(-1))\hat{k} = 3\hat{i} + 2\hat{j} - \hat{k}$.$\vec{BC} = \vec{c} - \vec{b} = (-13-5)\hat{i} + (-11-1)\hat{j} + (4-(-2))\hat{k} = -18\hat{i} - 12\hat{j} + 6\hat{k}$.$\vec{AC} = \vec{c} - \vec{a} = (-13-2)\hat{i} + (-11-(-1))\hat{j} + (4-(-1))\hat{k} = -15\hat{i} - 10\hat{j} + 5\hat{k}$.$\vec{CB} = \vec{b} - \vec{c} = (5-(-13))\hat{i} + (1-(-11))\hat{j} + (-2-4)\hat{k} = 18\hat{i} + 12\hat{j} - 6\hat{k}$.Now,solve for $\lambda$ in $\vec{AB} = \lambda \vec{BC}$:$3\hat{i} + 2\hat{j} - \hat{k} = \lambda (-18\hat{i} - 12\hat{j} + 6\hat{k}) = -6\lambda (3\hat{i} + 2\hat{j} - \hat{k})$.Thus,$-6\lambda = 1$,which gives $\lambda = -\frac{1}{6}$.Next,solve for $\mu$ in $\vec{AC} = \mu \vec{CB}$:$-15\hat{i} - 10\hat{j} + 5\hat{k} = \mu (18\hat{i} + 12\hat{j} - 6\hat{k})$.Dividing the components: $-15 = 18\mu \implies \mu = -\frac{15}{18} = -\frac{5}{6}$.Finally,$\lambda + \mu = -\frac{1}{6} + (-\frac{5}{6}) = -\frac{6}{6} = -1$.

Let $\vec{a} = 2\hat{i}-\hat{j}-\hat{k}$,$\vec{b} = 5\hat{i}+\hat{j}-2\hat{k}$,and $\vec{c} = -13\hat{i}-11\hat{j}+4\hat{k}$ be the position vectors of three points $A$,$B$,and $C$ respectively. If $\vec{AB} = \lambda \vec{BC}$ and $\vec{AC} = \mu \vec{CB}$,then find the value of $\lambda + \mu$.

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