If the points with position vectors hat{i}-2 | vedclass

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(B) Let $A, B, C$,and $D$ be the points with position vectors $\vec{A} = \hat{i}-2 \hat{j}+3 \hat{k}$,$\vec{B} = 2 \hat{i}+3 \hat{j}-4 \hat{k}$,$\vec{

Vedclass · Accepted Answer

$\frac{42}{19}$

Vedclass · Answer

(B) Let $A, B, C$,and $D$ be the points with position vectors $\vec{A} = \hat{i}-2 \hat{j}+3 \hat{k}$,$\vec{B} = 2 \hat{i}+3 \hat{j}-4 \hat{k}$,$\vec{C} = -3 \hat{i}+\hat{j}-5 \hat{k}$,and $\vec{D} = a \hat{i}-2 \hat{j}+4 \hat{k}$.The vectors formed are:$\overrightarrow{AB} = \vec{B} - \vec{A} = (2-1)\hat{i} + (3-(-2))\hat{j} + (-4-3)\hat{k} = \hat{i} + 5\hat{j} - 7\hat{k}$$\overrightarrow{AC} = \vec{C} - \vec{A} = (-3-1)\hat{i} + (1-(-2))\hat{j} + (-5-3)\hat{k} = -4\hat{i} + 3\hat{j} - 8\hat{k}$$\overrightarrow{AD} = \vec{D} - \vec{A} = (a-1)\hat{i} + (-2-(-2))\hat{j} + (4-3)\hat{k} = (a-1)\hat{i} + 0\hat{j} + \hat{k}$Since the points are coplanar,the scalar triple product of these vectors must be zero:$[\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}] = 0 \Rightarrow \begin{vmatrix} 1 & 5 & -7 \ -4 & 3 & -8 \ a-1 & 0 & 1 \end{vmatrix} = 0$Expanding along the third row:$(a-1) \begin{vmatrix} 5 & -7 \ 3 & -8 \end{vmatrix} - 0 \begin{vmatrix} 1 & -7 \ -4 & -8 \end{vmatrix} + 1 \begin{vmatrix} 1 & 5 \ -4 & 3 \end{vmatrix} = 0$$(a-1)(-40 - (-21)) + 1(3 - (-20)) = 0$$(a-1)(-19) + 23 = 0$$-19a + 19 + 23 = 0$$-19a + 42 = 0$$19a = 42 \Rightarrow a = \frac{42}{19}$

If the points with position vectors $\hat{i}-2 \hat{j}+3 \hat{k}$,$2 \hat{i}+3 \hat{j}-4 \hat{k}$,$-3 \hat{i}+\hat{j}-5 \hat{k}$,and $a \hat{i}-2 \hat{j}+4 \hat{k}$ are coplanar,then $a=$

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