The point if intersection of the lines l₁: r(t)=({i}-6{j}+2{k})+t({i}+2{j}+{k}) l₂: R(u)=(4{j}+{k})+u(2{i}+{j}+2{k}) { is}

Correct Answer is : (8,8,9)(8,8,9)(8,8,9)

Correct Answer is : (8,8,9)(8,8,9)(8,8,9)

Test Index

AP EAMCET Engineering 2012 Solved Paper

Section: Mathematics

Question : 154 of 160

Marks: +1, -0

l_{1}: r(t)=(\hat{i}-6\hat{j}+2\hat{k})+t(\hat{i}+2\hat{j}+\hat{k})

l_{2}: R(u)=(4\hat{j}+\hat{k})+u(2\hat{i}+\hat{j}+2\hat{k}) \text{ is}

Solution:

Put

r=x\hat{i}+y\hat{j}+z\hat{k}

\therefore x\hat{i}+y\hat{j}+z\hat{k}=(\hat{i}-6\hat{j}+2\hat{k})

+t(\hat{i}+2\hat{j}+\hat{k})

\therefore

Any point on line is

P(1+t,-6+2 t, 2+t)

is satisfied the second equation of line.

\therefore (1+t)\hat{i}+(-6+2 t)\hat{j}+(2+t)\hat{k}

=2 u\hat{i}+(4+u)\hat{j}+(1+2 u)\hat{k}

On equating the coefficients of

\hat{i}, \hat{j}

and

\hat{k}

, we get

1+t=2 u

⇒

t-2 u+1=0

........(i)

-6+2 t=4+u

⇒

2 t-u-10=0

........(ii)

and

2+t=1+2 u

⇒

t-2 u+1=0

.........(iii)

On solving Eqs. (i) and (ii), we get

t=7, u=4

\therefore P(1+7,-6+2 \times 7,2+7)

=P(8,8,9)

Go to Question:

Prev Question

Next Question