Let A, B, C be distinct points with position | vedclass

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(D) The position vectors of points $A, B, C$ are $\vec{a} = \hat{i} + \hat{j}$,$\vec{b} = \hat{i} - \hat{j}$,and $\vec{c} = p\hat{i} - q\hat{j} + r\ha

Vedclass · Accepted Answer

$p=1, q=2, r=0$

Vedclass · Answer

(D) The position vectors of points $A, B, C$ are $\vec{a} = \hat{i} + \hat{j}$,$\vec{b} = \hat{i} - \hat{j}$,and $\vec{c} = p\hat{i} - q\hat{j} + r\hat{k}$.For points $A, B, C$ to be collinear,the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ must be parallel,i.e.,$\overrightarrow{AC} = \lambda \overrightarrow{AB}$ for some scalar $\lambda$.Calculate $\overrightarrow{AB} = \vec{b} - \vec{a} = (\hat{i} - \hat{j}) - (\hat{i} + \hat{j}) = -2\hat{j}$.Calculate $\overrightarrow{AC} = \vec{c} - \vec{a} = (p-1)\hat{i} - (q+1)\hat{j} + r\hat{k}$.Since $\overrightarrow{AC} = \lambda \overrightarrow{AB}$,we have $(p-1)\hat{i} - (q+1)\hat{j} + r\hat{k} = \lambda(-2\hat{j}) = -2\lambda\hat{j}$.Comparing the coefficients of $\hat{i}, \hat{j}, \hat{k}$:$1$) $p-1 = 0 \implies p = 1$.$2$) $-(q+1) = -2\lambda \implies q+1 = 2\lambda$.$3$) $r = 0$.From option $D$,if $p=1, q=2, r=0$,then $2+1 = 2\lambda \implies 3 = 2\lambda \implies \lambda = 1.5$. This is a valid scalar,so the points are collinear. Thus,$p=1, q=2, r=0$ is a correct possibility.

Let $A, B, C$ be distinct points with position vectors $\hat{i} + \hat{j}$,$\hat{i} - \hat{j}$,and $p\hat{i} - q\hat{j} + r\hat{k}$ respectively. If points $A, B, C$ are collinear,then which of the following can be correct?

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