Sankalp JEE Full Test-1 Question-3 Solution

Question 3. Let the image of the point katex[/katex] in the line \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} be the point \left(\alpha,\;\beta,\;\gamma\right) Then which one of the following points lies on the line passing through \left(\alpha,\;\beta,\;\gamma\right) and making angles \frac{2\pi}{3} and \frac{3\pi}{4} with the y-axis and z-axis respectively, and an acute angle with the x-axis?

(1)\left(1,-2,1+\sqrt2\right)

(2) \left(1,2,1-\sqrt2\right)

(3) (3,4,3-2 \sqrt{2})

(4) (3,-4,3+2 \sqrt{2})

Answer (3)

Explanation:

L_{1}=\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda

M(\lambda, 1+2 \lambda, 2+3 \lambda) \overrightarrow{PM}=(\lambda-1) \hat{i}+(1+2 \lambda) \hat{j}+(3 \lambda-5) \hat{k}

\overrightarrow{PM} is perpendicular to line L_{1}

\overrightarrow{PM} \cdot \overrightarrow{b}=0 \quad(\overrightarrow{~b}=\hat{i}+2 \hat{j}+3 \hat{k}) \Rightarrow \lambda-1+4 \lambda+2+9 \lambda-15=0 14 \lambda=14 \Rightarrow \lambda=1 \therefore M=(1,3,5)

\overrightarrow{Q}=2 \overrightarrow{M}-\overrightarrow{P}[M is midpoint of \overrightarrow{P} \& \overrightarrow{Q}]

\overrightarrow{Q}=2 \hat{i}+6 \hat{j}+10 \hat{k}-\hat{i}-7 \hat{k} \overrightarrow{Q}=\hat{i}+6 \hat{j}+3 \hat{k} \therefore(\alpha, \beta, \gamma)=(1,6,3)

Required line having direction cosine ( l, ~m, ~n )
l^{2}+m^{2}+n^{2}=1

\Rightarrow l^{2}+\left(-\frac{1}{2}\right)^{2}+\left(-\frac{1}{\sqrt{2}}\right)^{2}=1 l^{2}=\frac{1}{4}

\therefore l=\frac{1}{2} [Line make acute angle with x -axis]

Equation of line passing through

will be \vec{r}=(\hat{i}+6 \hat{j}+3 \hat{k})+\mu\left(\frac{1}{2} \hat{i}-\frac{1}{2} \hat{j}-\frac{1}{\sqrt{2}} \hat{k}\right)

Option (3) satisfying for \mu=4