Find the value of $\lambda$ so that lines $\dfrac{1-x}{3} = \dfrac{7y-14}{2\lambda} = \dfrac{z-3}{2}$ and $\dfrac{7-7x}{3\lambda} = \dfrac{y-5}{1} = \dfrac{6-z}{5}$ are at right angle.

Find the value of λ so that lines (1-x)/(3) = (7y-14)/(2λ) = (z-3)/(2) and (7-7x)/(3λ) = (y-5)/(1) = (6-z)/(5)

Miscellaneous Exercise 6A · Q31

\lambda

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★

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Rewrite $\dfrac{1-x}{3}=\dfrac{7y-14}{2\lambda}=\dfrac{z-3}{2}$ in standard form. Since $1-x=-(x-1)$ , the first fraction is $\dfrac{x-1}{-3}$ . Since $7y-14=7(y-2)$ , the second fraction is $\dfrac{y-2}{2\lambda/7}$ . So this line has direction ratios

d_1=\left(-3,\ \dfrac{2\lambda}{7},\ 2\right).

Similarly, $\dfrac{7-7x}{3\lambda}=\dfrac{y-5}{1}=\dfrac{6-z}{5}$ : since $7-7x=-7(x-1)$ , the first fraction is $\dfrac{x-1}{-3\lambda/7}$ ; since $6-z=-(z-6)$ , the third fraction is $\dfrac{z-6}{-5}$ . So this line has direction ratios

$d_2=\left(-\dfrac{3\lambda}{7},\ 1,\ -5\right).$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2018Set ANNUAL2 marksMCQ
$-2$
(a) $\cos^{-1}\left(\dfrac{1}{7}\right)$
(b) $\cos^{-1}\left(\dfrac{8}{15}\right)$
(c) $\cos^{-1}\left(\dfrac{1}{3}\right)$
(d) $\cos^{-1}\left(\dfrac{8}{21}\right)$
›Reveal solutionSolution

Use $\cos\theta = \dfrac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$ .

Direction ratios: $(a_1,b_1,c_1) = (3,2,6)$ and $(a_2,b_2,c_2) = (-2,1,2)$ .

$\cos\theta = \frac{3(-2)+2(1)+6(2)}{\sqrt{3^2+2^2+6^2}\,\sqrt{(-2)^2+1^2+2^2}} = \frac{-6+2+12}{\sqrt{9+4+36}\,\sqrt{4+1+4}} = \frac{8}{\sqrt{49}\sqrt{9}} = \frac{8}{7\times3} = \frac{8}{21}$

…

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CBSE 2016Set ANNUAL2 marks
$\dfrac{x-1}{-3} = \dfrac{y-2}{2k} = \dfrac{z-3}{2}$
›Reveal solutionSolution

Two lines are at right angles when the dot product of their direction ratios is zero.

Direction ratios of the lines:

Line 1: $(-3,\,2k,\,2)$

Line 2: $(3k,\,1,\,-5)$

For lines at right angles, the dot product of direction ratios must be zero:

…

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Q.Find the value of $\lambda$ so that lines $\dfrac{1-x}{3} = \dfrac{7y-14}{2\lambda} = \dfrac{z-3}{2}$ and $\dfrac{7-7x}{3\lambda} = \dfrac{y-5}{1} = \dfrac{6-z}{5}$ are at right angle.

Concept understanding — Angle between Two Lines

Q.Find the value of λ\lambdaλ so that lines 1−x3=7y−142λ=z−32\dfrac{1-x}{3} = \dfrac{7y-14}{2\lambda} = \dfrac{z-3}{2}31−x​=2λ7y−14​=2z−3​ and 7−7x3λ=y−51=6−z5\dfrac{7-7x}{3\lambda} = \dfrac{y-5}{1} = \dfrac{6-z}{5}3λ7−7x​=1y−5​=56−z​ are at right angle.

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Q.Find the value of $\lambda$ so that lines $\dfrac{1-x}{3} = \dfrac{7y-14}{2\lambda} = \dfrac{z-3}{2}$ and $\dfrac{7-7x}{3\lambda} = \dfrac{y-5}{1} = \dfrac{6-z}{5}$ are at right angle.

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