The lines overline{r}=overline{a}+lambda(over | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Two lines $\overline{r} = \overline{a_1} + \lambda \overline{v_1}$ and $\overline{r} = \overline{a_2} + \mu \overline{v_2}$ intersect if and only

Vedclass · Accepted Answer

$[\overline{a} \overline{b} \overline{c}] = 0$

Vedclass · Answer

(B) Two lines $\overline{r} = \overline{a_1} + \lambda \overline{v_1}$ and $\overline{r} = \overline{a_2} + \mu \overline{v_2}$ intersect if and only if the shortest distance between them is $0$. The condition for intersection is $(\overline{a_2} - \overline{a_1}) \cdot (\overline{v_1} 	imes \overline{v_2}) = 0$. Here,$\overline{a_1} = \overline{a}$,$\overline{v_1} = \overline{b} 	imes \overline{c}$,$\overline{a_2} = \overline{c}$,and $\overline{v_2} = \overline{a} 	imes \overline{b}$. Substituting these into the condition: $(\overline{c} - \overline{a}) \cdot ((\overline{b} 	imes \overline{c}) 	imes (\overline{a} 	imes \overline{b})) = 0$. Using the vector triple product identity $(\overline{x} 	imes \overline{y}) 	imes \overline{z} = (\overline{x} \cdot \overline{z})\overline{y} - (\overline{y} \cdot \overline{z})\overline{x}$,we simplify $(\overline{b} 	imes \overline{c}) 	imes (\overline{a} 	imes \overline{b})$. Let $\overline{d} = \overline{a} 	imes \overline{b}$. Then $(\overline{b} 	imes \overline{c}) 	imes \overline{d} = (\overline{b} \cdot \overline{d})\overline{c} - (\overline{c} \cdot \overline{d})\overline{b}$. Since $\overline{d} = \overline{a} 	imes \overline{b}$,$\overline{b} \cdot \overline{d} = 0$ and $\overline{c} \cdot \overline{d} = [\overline{c} \overline{a} \overline{b}] = [\overline{a} \overline{b} \overline{c}]$. Thus,the vector product is $-[\overline{a} \overline{b} \overline{c}] \overline{b}$. The condition becomes $(\overline{c} - \overline{a}) \cdot (-[\overline{a} \overline{b} \overline{c}] \overline{b}) = 0$. This implies $-[\overline{a} \overline{b} \overline{c}] (\overline{c} \cdot \overline{b} - \overline{a} \cdot \overline{b}) = 0$. This is satisfied if $[\overline{a} \overline{b} \overline{c}] = 0$.

The lines $\overline{r}=\overline{a}+\lambda(\overline{b} \times \overline{c})$ and $\overline{r}=\overline{c}+\mu(\overline{a} \times \overline{b})$ will intersect if

Similar Questions

The altitude of the parallelepiped,whose coterminus edges are the vectors $\bar{a}=\hat{i}+\hat{j}+\hat{k}$,$\bar{b}=2\hat{i}+4\hat{j}-\hat{k}$,and $\bar{c}=\hat{i}+\hat{j}+3\hat{k}$,where $\bar{a}$ and $\bar{b}$ are the sides of the base of the parallelepiped,is:

If the vectors $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+2\hat{k}$ and $\vec{c}=x\hat{i}+(x-2)\hat{j}-\hat{k}$ are coplanar, then $x=$

Let $a, b, c$ be three distinct real numbers,none equal to $1$. If the vectors $a \hat{i}+\hat{j}+\hat{k}$,$\hat{i}+b \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+ c \hat{k}$ are coplanar,then $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module