The locus of the point P(vec{r}) which forms | vedclass

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(C) Let the point $P(\vec{r})$ be $x\hat{i}+y\hat{j}+z\hat{k}$. The fixed points are $A(1, 0, 0)$ and $B(0, 1, 0)$.Vector $\vec{AP} = (x-1)\hat{i} + y

Vedclass · Accepted Answer

$(x+y-1)^2+2z^2=4$

Vedclass · Answer

(C) Let the point $P(\vec{r})$ be $x\hat{i}+y\hat{j}+z\hat{k}$. The fixed points are $A(1, 0, 0)$ and $B(0, 1, 0)$.Vector $\vec{AP} = (x-1)\hat{i} + y\hat{j} + z\hat{k}$ and vector $\vec{AB} = -\hat{i} + \hat{j}$.The area of $	riangle ABP$ is given by $\frac{1}{2} |\vec{AP} 	imes \vec{AB}|$.Calculating the cross product:$\vec{AP} 	imes \vec{AB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ x-1 & y & z \ -1 & 1 & 0 \end{vmatrix} = \hat{i}(0-z) - \hat{j}(0 - (-z)) + \hat{k}((x-1) - (-y)) = -z\hat{i} - z\hat{j} + (x+y-1)\hat{k}$.The magnitude is $|\vec{AP} 	imes \vec{AB}| = \sqrt{(-z)^2 + (-z)^2 + (x+y-1)^2} = \sqrt{2z^2 + (x+y-1)^2}$.Given the area is $1$, we have $1 = \frac{1}{2} \sqrt{2z^2 + (x+y-1)^2}$.Squaring both sides: $2 = \sqrt{2z^2 + (x+y-1)^2} \Rightarrow 4 = 2z^2 + (x+y-1)^2$.

The locus of the point $P(\vec{r})$ which forms a triangle $ABP$ of area $1$ sq. unit with the fixed points $A(\hat{i})$ and $B(\hat{j})$ is

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