- Column I Column II- a- Lines x-1-2 - y -23 - z-1 and- r - 3- - - - k- - t- - - - k- are p- intersecting- b- Lines x-51 - y-37 - z-33 and- x - y - 2z - 4 - 0 - 2x - y - 3z - 5 - 0- are q- perpendicular- c- Lines x - t - 3- y- -2t - 1-z - -3t - 2- and r - t - 1- - 2t - 3- - -t - 9 k- are r- parallel- d- Lines r - - - 3- - k-t2- - - - k- and- r - - - 2- - 5k- - s- - 2- - 34k- are s- skew- -Then which of the following is correct -

A. Line x 1 2 = y + 23 = z 1 nbsp; is along the vector a = 2î + 3ĵ k̂ and line r = (3î ĵ + k̂) + t(î + ĵ +k̂) is along the vector b = î + ĵ + k̂. Here a ...

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Byju's Answer

Standard XIII

Mathematics

Condition for Two Lines to Be Perpendicular

[ ...

Question

$\begin{matrix} Column I & Column II a. Lines \frac{x - 1}{- 2} = \frac{y + 2}{3} = \frac{z}{- 1} and \to r = (3^i -^j +^k) + t (^i +^j +^k) are & p. intersecting b. Lines \frac{x + 5}{1} = \frac{y - 3}{7} = \frac{z + 3}{3} and x - y + 2 z - 4 = 0 = 2 x + y - 3 z + 5 = 0 are & q. perpendicular c. Lines (x = t - 3, y = - 2 t + 1, z = - 3 t - 2) and \to r = (t + 1)^i + (2 t + 3)^j + (- t - 9)^k are & r. parallel d. Lines \to r = (^i + 3^j -^k) + t (2^i -^j -^k) and \to r = (-^i - 2^j + 5^k) + s (^i - 2^j + \frac{3}{4}^k) are & s. skew \end{matrix}$

Then which of the following is correct ?

a \to q, s; b \to r; c \to p, q; d \to p

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a \to q; b \to r; c \to p, q; d \to p, q

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a \to s; b \to r, s; c \to p; d \to p, q

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a \to q, s; b \to r, s; c \to p; d \to p

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Solution

The correct option is A $a \to q, s; b \to r; c \to p, q; d \to p$
$a .$
Line $\frac{x - 1}{- 2} = \frac{y + 2}{3} = \frac{z}{- 1}$ is along the vector $\to a = - 2^i + 3^j -^k$ and line $\to r = (3^i -^j +^k) + t (^i +^j +^k)$ is along the vector $\to b =^i +^j +^k .$
Here $\to a ⊥ \to b .$

Also, $∣ ∣ ∣ ∣ \begin{matrix} 3 - 1 & - 1 - (- 2) & 1 - 0 - 2 & 3 & - 1 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ \neq 0$

$b .$
The direction ratios of the line $x - y + 2 z - 4 = 0 = 2 x + y - 3 z + 5 = 0$ are $∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & - 1 & 2 2 & 1 & - 3 \end{matrix} ∣ ∣ ∣ ∣ ∣ =^i + 7^j + 3^k .$
Hence, the given two lines are parallel.

$c .$
The given lines are $(x = t - 3, y = - 2 t + 1, z = - 3 t - 2)$ and $\to r = (t + 1)^i + (2 t + 3)^j + (- t - 9)^k,$

$\Rightarrow \frac{x + 3}{1} = \frac{y - 1}{- 2} = \frac{z + 2}{- 3}$
and $\frac{x - 1}{1} = \frac{y - 3}{2} = \frac{z + 9}{- 1}$

The lines are perpendicular as $(1) (1) + (- 2) (2) + (- 3) (- 1) = 0.$
Also, $∣ ∣ ∣ ∣ \begin{matrix} - 3 - 1 & 1 - 3 & - 2 - (- 9) 1 & - 2 & - 3 1 & 2 & - 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$

Hence, the lines are intersecting.

$d .$
The given lines are $\to r = (^i + 3^j -^k) + t (2^i -^j -^k)$ and $\to r = (-^i - 2^j + 5^k) + s (^i - 2^j + \frac{3}{4}^k) .$
$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 - (- 1) & 3 - (- 2) & - 1 - 5 2 & - 1 & - 1 1 & - 2 & 3 / 4 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$

Hence, the lines are coplanar and intersecting (as the lines are not parallel).

Suggest Corrections

Similar questions

\begin{matrix} Column I & Column II a. Lines \frac{x - 1}{- 2} = \frac{y + 2}{3} = \frac{z}{- 1} and \to r = (3^i -^j +^k) + t (^i +^j +^k) are & p. intersecting b. Lines \frac{x + 5}{1} = \frac{y - 3}{7} = \frac{z + 3}{3} and x - y + 2 z - 4 = 0 = 2 x + y - 3 z + 5 = 0 are & q. perpendicular c. Lines (x = t - 3, y = - 2 t + 1, z = - 3 t - 2) and \to r = (t + 1)^i + (2 t + 3)^j + (- t - 9)^k are & r. parallel d. Lines \to r = (^i + 3^j -^k) + t (2^i -^j -^k) and \to r = (-^i - 2^j + 5^k) + s (^i - 2^j + \frac{3}{4}^k) are & s. skew \end{matrix}

Then which of the following is correct ?

\begin{matrix} Column I & Column II a & The distance between the line \to r = (2^i - 2^j + 3^k) + λ (^i -^j + 4^k) & p & \frac{25}{3 \sqrt{14}} and plane \to r . (^i + 5^j +^k) = 5 b & The distance between parallel planes \to r . (2^i -^j + 3^k) = 4 and & q & \frac{13}{7} \to r . (6^i - 3^j + 9^k) + 13 = 0 is c & The distance of a point (2, 5, -3) from the plane \to r . (6^i - 3^j + 2^k) = 4 & r & \frac{10}{3 \sqrt{3}} is d & The distance of the point (1, 0, -3) from the plane x - y - z - 9 = 0 & s & 7 measured parallel to line \frac{x - 2}{2} = \frac{y + 2}{3} = \frac{z - 6}{- 6} \end{matrix}

Then which of the following is correct ?

Q. Let

L_{1}

, and

L_{2}

denotes the lines

\to r =^i + λ (-^i + 2^j + 2^k), λ \in R

and

\to r = μ (2^i -^j + 2^k), μ \in R

respectively.
If

L_{3}

is a line which is perpendicular to both

L_{1}

and

L_{2}

and cuts both of them, then which of the following option describe(s)

L_{3}

Equation of the plane containing the lines $\to r =^i +^j -^k + λ (^i + 2^j -^k) a n d \to r =^i + 2^j -^k + μ (^i +^j + 3^k)$ is

Q. Match

List I

with the

List II

and select the correct answer using the code given below the lists :

\begin{matrix} List I & List II (A) & If \to a = t^i - 3^j + 2 t^k, \to b =^i - 2^j + 2^k and \to c = 3^i + t^j -^k, then the & (P) & 1 value of ⎛ ⎜ ⎝ 2 \int 1 \to a \cdot (\to b \times \to c) d t + 2 ⎞ ⎟ ⎠ is equal to (B) & The value of t \in R for which the vectors \to a = (1, - 2, 3), \to b = (- 2, 3, - 4), & (Q) & 2 \to c = (1, - 1, t) form a linearly dependent system is (C) & If the area of the parallelogram whose diagonals are 3^i +^j - 2^k and & (R) & 3^i - 3^j + 4^k is μ \sqrt{3} sq. unit, then the value of μ is (D) & Let \to p =^i +^j, \to q =^i -^j and \to r =^i + 2^j + 3^k . If \to x is a unit vector & (S) & 4 such that \to p \cdot \to x = 0 and \to q \cdot \to x = 0, then | \to r \cdot \to x | is equal to (T) & 5 \end{matrix}

Which of the following is the only CORRECT combination ?

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