Question

The direction ratios of a line \(L\) are \((ab,b,b)\) \((b\gt 0)\) and the line \(L\) passes through \[ P(b,b,b). \] If \(Q(x,y,z)\) is a point on the line at a distance of \(b\) units from \(P\), then \[ x+y+z= \]

• A. \[ b\left(3\pm\frac{a+2}{\sqrt{a^2+2}}\right) \]
• B. \[ b\left(3\pm\frac{a^2-2}{\sqrt{a^2+2}}\right) \]
• C. \[ b\left(a\pm\frac{a+2}{\sqrt{a^2+2}}\right) \]
• D. \[ b\left(a\pm\frac{a^2-2}{\sqrt{a^2+2}}\right) \]

Hint:

To find a point at a given distance along a line, first convert the direction ratios into a unit vector. Then use \[ \text{New Point} = \text{Given Point} \pm (\text{Distance})\times(\text{Unit Vector}). \]

Answer 1

The Correct Option is A

Concept: If a line passes through a point \[ (x_1,y_1,z_1) \] and has direction ratios \[ (l,m,n), \] then a point at distance \(d\) from the given point along the line is obtained by moving \(d\) units in the direction of the unit vector of \((l,m,n)\).

Step 1: Find the unit vector along the line. Direction ratios are \[ (ab,b,b). \] Its magnitude is \[ \sqrt{(ab)^2+b^2+b^2} = b\sqrt{a^2+2}. \] Since \(b\gt 0\), \[ \hat{u} = \left( \frac{a}{\sqrt{a^2+2}}, \frac{1}{\sqrt{a^2+2}}, \frac{1}{\sqrt{a^2+2}} \right). \]

Step 2: Find the coordinates of \(Q\). Given \[ P=(b,b,b) \] and \[ PQ=b. \] Therefore, \[ Q = P \pm b\hat{u}. \] Hence, \[ x = b\pm\frac{ab}{\sqrt{a^2+2}}, \] \[ y = b\pm\frac{b}{\sqrt{a^2+2}}, \] \[ z = b\pm\frac{b}{\sqrt{a^2+2}}. \]

Step 3: Compute \(x+y+z\). Adding, \[ x+y+z = 3b \pm \frac{ab+b+b}{\sqrt{a^2+2}}. \] \[ = 3b \pm \frac{b(a+2)}{\sqrt{a^2+2}}. \] \[ = b\left( 3\pm\frac{a+2}{\sqrt{a^2+2}} \right). \]

Step 4: Write the final answer. \[ \boxed{ b\left( 3\pm\frac{a+2}{\sqrt{a^2+2}} \right) } \]