The position vectors of the points A and B ar | vedclass

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(A) Given the position vectors of $A$ and $B$ are $A(1, 2, 0)$ and $B(2, 1, 1)$.The equation of the plane is $x+y+z=3$.The normal vector to the plane

Vedclass · Accepted Answer

$\frac{2 \sqrt{2}}{\sqrt{3}}$

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(A) Given the position vectors of $A$ and $B$ are $A(1, 2, 0)$ and $B(2, 1, 1)$.The equation of the plane is $x+y+z=3$.The normal vector to the plane is $\vec{n} = \hat{i}+\hat{j}+\hat{k}$.The line passing through $A$ and perpendicular to the plane is given by $\vec{r} = (\hat{i}+2\hat{j}) + \lambda(\hat{i}+\hat{j}+\hat{k})$.Any point on this line is $(1+\lambda, 2+\lambda, \lambda)$. Since this point lies on the plane $x+y+z=3$, we have $(1+\lambda) + (2+\lambda) + \lambda = 3$, which gives $3\lambda + 3 = 3$, so $\lambda = 0$. Thus, $P = (1, 2, 0)$.The line passing through $B$ and perpendicular to the plane is given by $\vec{r} = (2\hat{i}+\hat{j}+\hat{k}) + \mu(\hat{i}+\hat{j}+\hat{k})$.Any point on this line is $(2+\mu, 1+\mu, 1+\mu)$. Since this point lies on the plane $x+y+z=3$, we have $(2+\mu) + (1+\mu) + (1+\mu) = 3$, which gives $3\mu + 4 = 3$, so $\mu = -\frac{1}{3}$.Thus, $Q = (2-\frac{1}{3}, 1-\frac{1}{3}, 1-\frac{1}{3}) = (\frac{5}{3}, \frac{2}{3}, \frac{2}{3})$.The distance $PQ$ is given by $\sqrt{(\frac{5}{3}-1)^2 + (\frac{2}{3}-2)^2 + (\frac{2}{3}-0)^2} = \sqrt{(\frac{2}{3})^2 + (-\frac{4}{3})^2 + (\frac{2}{3})^2} = \sqrt{\frac{4}{9} + \frac{16}{9} + \frac{4}{9}} = \sqrt{\frac{24}{9}} = \frac{\sqrt{24}}{3} = \frac{2\sqrt{6}}{3} = \frac{2\sqrt{2}}{\sqrt{3}}$.

The position vectors of the points $A$ and $B$ are respectively $\hat{i}+2 \hat{j}$ and $2 \hat{i}+\hat{j}+\hat{k}$. If the points $P$ and $Q$ are respectively the orthogonal projections of $A$ and $B$ on the plane $x+y+z=3$, then $P Q=$

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