Single Choice

The shortest distance between the $$z$$-axis and the line $$x+ y + 2z - 3 = 0 , 2x + 3y + 4z - 4 $$, is

A$$1$$
B$$2$$
Correct Answer
C$$3$$
D$$4$$

Solution

Any two point on the line $$x+ y + 2z - 3 = 0 , 2x + 3y + 4z - 4 $$ are $$(1,-2,2)$$ and $$(5,-2,0)$$

Thus its equation may be written as, $$\displaystyle \frac{x-5}{4}=\frac{y+2}{0}=\frac{z}{-2}$$

and equation of line $$z$$-axis is, $$\displaystyle \frac{x}{0}=\frac{y}{0}=\frac{z}{1}$$

Now we know shortest distance between two line is given by, $$\displaystyle D = \frac{(\vec{a}-\vec{c}).(\vec{b}\times \vec{d})}{|\vec{b}\times \vec{d}|}$$

Here $$\vec{a} = 0, \vec{d} = 4i-2k,\vec{b} = k,\vec{c} =5i-2j$$

$$\therefore \displaystyle \vec{b}\times\vec{d}=\begin{vmatrix} \hat i&\hat j&\hat k\\ 0&0&1\\4&0&-2\end{vmatrix}=4\hat j$$

Thus $$\displaystyle D = \frac{(5\hat i-2\hat j).(4\hat j)}{|4\hat j|}=\frac{8}{4}=2$$

SIMILAR QUESTIONS

3D Geometry

Consider the line $$\displaystyle {L}_{1}:\dfrac{{x}+1}{3}=\dfrac{{y}+2}{1}=\dfrac{{z}+1}{2},\ \displaystyle {L}_{2}:\dfrac{{x}-2}{1}=\dfrac{{y}+2}{2}=\dfrac{{z}-3}{3}$$ The shortest distance between $$L_1$$ and $$L_2$$ is

3D Geometry

If the shortest distance between the lines $$\displaystyle \frac{x-1}{\alpha} = \frac{y+1}{-1} = \frac{z}{1}, \,\,\, (\alpha \neq -1)$$ and $$x + y + z + 1 = 0 = 2x - y + z + 3$$ is $$\displaystyle \frac{1}{\sqrt{3}}$$, then a value of $$\alpha$$ is :

3D Geometry

Find the shortest distance between the following pair of lines. $$\bar{r}=(\bar{i}+2\bar{j}+\bar{k})+\lambda(2\bar{i}-\bar{j}+3\bar{k})$$ & $$\bar{r}=(\bar{i}-3\bar{j}-\bar{k})+\mu (3\bar{i}+2\hat{j}-5\bar{k})$$.

3D Geometry

The shortest distance between the lines $$\dfrac {x}{2} = \dfrac {y}{2} = \dfrac {z}{1}$$ and $$\dfrac {x + 2}{-1} = \dfrac {y - 4}{8} = \dfrac {z - 5}{4}$$ lies in the interval:

3D Geometry

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3D Geometry

Find the shortest distance between the lines $$\vec { r } =\left( \hat { i } +2\hat { j } +\hat { k } \right) +\lambda \left( \hat { i } -\hat { j } +\hat { k } \right) $$ and $$\vec { r } =\hat { 2i } -\hat { j } -\hat { k } +\mu \left( \hat { 2i } +\hat { j } +2\hat { k } \right) $$

3D Geometry

Find the shortest distance between lines $$\cfrac{x+1}{7}=\cfrac{y+1}{-6}=\cfrac{z+1}{1}$$ and $$\cfrac{x-3}{1}=\cfrac{y-5}{-2}=\cfrac{z-7}{1}$$

3D Geometry

Find the shortest distance between the lines whose vector equations are $$\vec {r} = (\hat {i} + 2\hat {j} + 3\hat {k}) + \lambda (\hat {i} - 3\hat {j} + 2\hat {k})$$ and $$\vec {r} = (4\hat {i} + 5\hat {j} + 6\hat {k}) + \mu (2\hat {i} + 3\hat {j} + \hat {k})$$.

3D Geometry

Find the shortest distance between the lines whose vector equations are : $$\vec { r } =(1-p)\hat { i } +(p-2)\hat { j } +(3-2p)\hat { k } $$ $$\vec { r } =(q+1)\hat { i } +(2q-1)\hat { j } -(2q-1)\hat { k } $$

3D Geometry

Find the shortest distance between lines $$\vec { r } =6\hat { i } +2\hat { j } +2\hat { k } +\lambda \left( \hat { i } -2\hat { j } +2\hat { k } \right) $$ and $$\vec { r } =-4\hat { i } -\hat { k } +\mu \left(3 \hat { i } -2\hat { j } -2\hat { k } \right) $$

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