Let vec{a}, vec{b}, vec{c} be three non-copla | vedclass

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(C) Let the points be $A = \vec{a}+\vec{b}+\vec{c}$,$B = \vec{a}-\vec{b}+3\vec{c}$,$C = 2\vec{a}-\vec{b}+\vec{c}$,and $D = \vec{a}-2\vec{b}+4\vec{c}$.

Vedclass · Accepted Answer

$\vec{a}-2 \vec{b}+4 \vec{c}$

Vedclass · Answer

(C) Let the points be $A = \vec{a}+\vec{b}+\vec{c}$,$B = \vec{a}-\vec{b}+3\vec{c}$,$C = 2\vec{a}-\vec{b}+\vec{c}$,and $D = \vec{a}-2\vec{b}+4\vec{c}$.The line passing through $A$ and $B$ is given by $\vec{r} = A + \lambda_1(B-A) = (\vec{a}+\vec{b}+\vec{c}) + \lambda_1(-2\vec{b}+2\vec{c}) = \vec{a} + (1-2\lambda_1)\vec{b} + (1+2\lambda_1)\vec{c}$.The line passing through $C$ and $D$ is given by $\vec{r} = C + \lambda_2(D-C) = (2\vec{a}-\vec{b}+\vec{c}) + \lambda_2(-\vec{a}-\vec{b}+3\vec{c}) = (2-\lambda_2)\vec{a} + (-1-\lambda_2)\vec{b} + (1+3\lambda_2)\vec{c}$.Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar,we equate the coefficients of $\vec{a}, \vec{b}, \vec{c}$:$1 = 2-\lambda_2 \implies \lambda_2 = 1$.$1-2\lambda_1 = -1-\lambda_2 = -1-1 = -2 \implies 2\lambda_1 = 3 \implies \lambda_1 = \frac{3}{2}$.Check the coefficient of $\vec{c}$: $1+2\lambda_1 = 1+2(\frac{3}{2}) = 4$ and $1+3\lambda_2 = 1+3(1) = 4$. They match.Substituting $\lambda_1 = \frac{3}{2}$ into the first line equation:$\vec{r} = \vec{a} + (1-2(\frac{3}{2}))\vec{b} + (1+2(\frac{3}{2}))\vec{c} = \vec{a} - 2\vec{b} + 4\vec{c}$.

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors. Then the point of intersection of the line joining the points $\vec{a}+\vec{b}+\vec{c}, \vec{a}-\vec{b}+3 \vec{c}$ and the line joining the points $2 \vec{a}-\vec{b}+\vec{c}, \vec{a}-2 \vec{b}+4 \vec{c}$ is

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