Show that the lines r =3 i +2 j -4 k + ( i +2 j +2 k ) and r =5 i -2 j + (3 i +2 j +6 k ) are intersecting. Hence, find their point of intersection.

The position vectors of two arbitrary points on the given lines are 3 i +2 j -4 k + ( i +2 j +2 k ) =(3+ ) i +(2+2 ) j +(2 -4) k 5 i -2 j + (3 i +2 j +6 k ) =(5+3 ) i +(-2+2 ) j +6 k If the lines intersect, then they have a common point. so, for some values of and , we must have (3+ ) i +(2+2 ) j +(2 -4) k =(5+3 ) i +(-2+2 ) j 6 k Equation the coefficients of i , j and k , we get 3+ =5+3 (1) 2+2 =-2+2 (2) 2 -4=6 (3) Solving (1) and (2), we get =-4, =-2. Substituting the values =-4 and =-2 in (3), we get LHS=2 -4 =2(-4)-4 =-12 RHS=6 =6(-2) =-12 =RHS Since =-4 and =-2 satisfy (3), the lines intersect. Substituting =-2 in the second line, we get r =5 i -2 j -6 i -4 j -12 k =- i -6 j -12 k the position vector of the point of intersection. Thus, the coordinates of the points of intersection are (-1, -6, -12).

Show that the lines r =3 i +2 j -4 k + ( i +2 j +2 k ) and r =5 i…

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Question

Show that the lines
$\vec{\text{r}}=3\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)$ and $\vec{\text{r}}=5\hat{\text{i}}-2\hat{\text{j}}+\mu\big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\big)$ are intersecting. Hence, find their point of intersection.

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Answer

The position vectors of two arbitrary points on the given lines are
$3\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)$
$=(3+\lambda)\hat{\text{i}}+(2+2\lambda)\hat{\text{j}}+(2\lambda-4)\hat{\text{k}}$
$5\hat{\text{i}}-2\hat{\text{j}}+\mu\big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\big)$
$=(5+3\mu)\hat{\text{i}}+(-2+2\mu)\hat{\text{j}}+6\mu\hat{\text{k}}$
If the lines intersect, then they have a common point. so, for some values of $\lambda$ and $\mu,$ we must have
$(3+\lambda)\hat{\text{i}}+(2+2\lambda)\hat{\text{j}}+(2\lambda-4)\hat{\text{k}}$
$=(5+3\mu)\hat{\text{i}}+(-2+2\mu)\hat{\text{j}}6\mu\hat{\text{k}}$
Equation the coefficients of $\hat{\text{i}},\hat{\text{j}}$ and $\hat{\text{k}},$ we get
$3+\lambda=5+3\mu\dots(1)$
$2+2\lambda=-2+2\mu\dots(2)$
$2\lambda-4=6\mu\dots(3)$
Solving (1) and (2), we get
$\lambda=-4,\mu=-2.$
Substituting the values $\lambda=-4$ and $\mu=-2$ in (3), we get
$\text{LHS}=2\lambda-4$
$=2(-4)-4$
$=-12\text{ RHS}=6\mu$
$=6(-2)$
$=-12$
$\Rightarrow\text{LHS}=\text{RHS}$
Since $\lambda=-4$ and $\mu=-2$ satisfy (3), the lines intersect.
Substituting $\mu=-2$ in the second line, we get $\vec{\text{r}}=5\hat{\text{i}}-2\hat{\text{j}}-6\hat{\text{i}}-4\hat{\text{j}}-12\hat{\text{k}}=-\hat{\text{i}}-6\hat{\text{j}}-12\hat{\text{k}}$ the position vector of the point of intersection.
Thus, the coordinates of the points of intersection are (-1, -6, -12).

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