Assertion (A): Lines given by $x = py + q$ , $z = ry + s$ and $x = p'y + q'$ , $z = r'y + s'$ are perpendicular to each other when $pp' + rr' = 1$ . Reason (R): Two lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ are perpendicular to each other if $\vec{b}_1 \cdot \vec{b}_2 = 0$ .

Q: Assertion (A): Lines given by $x = py + q$, $z = ry + s$ and $x = p'y + q'$, $z = r'y + s'$ are perpendicular to each other when $pp' + rr' = 1$. Reason (R): Two lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ are perpendicular to each other if $\vec{b}_1 \cdot \vec{b}_2 = 0$.

The correct option is **(D)** — Assertion (A) is false, but Reason (R) is true. **Checking the Assertion:** Writing each line with y as the parameter gives direction vectors $(p, 1, r)$ and $(p', 1, r')$. Two lines are perpendicular when the dot product of their direction vectors is zero: $pp' + 1 + rr' = 0$, i.e. $pp' + rr' = -1$, NOT +1 as the Assertion claims. So the Assertion is false. **Checking the Reason:** The condition $b_1 \cdot b_2 = 0$ for perpendicularity of two lines in vector form is the standard, universally true definition.

Q: What is the correct condition for the two lines in Q20 to be perpendicular?

$pp' + rr' = -1$, obtained by setting the dot product of direction vectors $(p,1,r)$ and $(p',1,r')$ to zero: $pp' + 1 + rr' = 0$.

Q: Why is direction vector (p,1,r) used instead of (p,0,r)?

Because in the given line equations $x = py+q$, $z = ry+s$, the variable y itself is the natural parameter (coefficient 1), while x and z are expressed in terms of y — so the direction vector's y-component is 1, not 0.

Three Dimensional Geometry

Perpendicular Lines in 3D

Options

(A)Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).

(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

(C)Assertion (A) is true and Reason (R) is false.

(D)Assertion (A) is false and Reason (R) is true.

Official Answer

The correct option is (D) — Assertion (A) is false, but Reason (R) is true.

Checking the Assertion: Writing each line with y as the parameter gives direction vectors $(p, 1, r)$ and $(p', 1, r')$ . Two lines are perpendicular when the dot product of their direction vectors is zero: $pp' + 1 + rr' = 0$ , i.e. $pp' + rr' = -1$ , NOT +1 as the Assertion claims. So the Assertion is false.

Checking the Reason: The condition $b_1 \cdot b_2 = 0$ for perpendicularity of two lines in vector form is the standard, universally true definition.

direction vectorperpendicular linesdot productthree dimensional geometrysymmetric form of a lineassertion reason

Marking Scheme

11 mark: correct option (D) — full marks require correctly identifying the Assertion as false and the Reason as true; no partial credit.

Hint

Write each line's direction vector as $(p,1,r)$ type (treating y as the parameter) and set the dot product of the two direction vectors to zero.

Quick Oral Answer

Writing each line's direction vector as $(p,1,r)$ , perpendicularity needs $pp' + 1 + rr' = 0$ , so the correct condition is $pp'+rr' = -1$ , not +1 as the Assertion claims — so A is false, R ( $b_1 \cdot b_2 = 0$ ) is true.

Analysis & Explanation

This question checks whether students can correctly derive the perpendicularity condition for two lines given in the 'x = py+q, z = ry+s' symmetric form.

Deriving the correct condition: Treating y as the parameter, line 1 can be written as $(x,y,z) = (q,0,s) + y(p,1,r)$ , so its direction vector is $(p,1,r)$ . Similarly line 2's direction vector is $(p',1,r')$ . Two lines are perpendicular exactly when the dot product of their direction vectors is zero:

$p \cdot p' + 1 \cdot 1 + r \cdot r'$ = 0 ⟹ $pp' + rr' = -1$

Why (D) is correct: The Assertion states the condition as $pp' + rr' = 1$ , but the correct condition (derived above) is $pp' + rr' = -1$ . So the Assertion is false. The Reason, $b_1 \cdot b_2 = 0$ for perpendicular lines $r = a_1+\lambda b_1$ and $r = a_2+\mu b_2$ , is the standard vector-form perpendicularity condition and is true — it is in fact the very principle used above to show the Assertion is wrong.

Why the distractors are wrong:

(A) and (B) require the Assertion to be true, but the sign is wrong (should be −1, not 1).
(C) requires the Reason to be false, but b₁·b₂=0 is a correct, standard definition.

Common Mistakes

1Forgetting that y is the natural parameter in this form and mis-identifying the direction ratios as $(p,0,r)$ instead of $(p,1,r)$ .
2Sign error when equating the dot product to zero, concluding $pp'+rr' = 1$ instead of −1.
3Assuming the Reason is false simply because it looks like a 'different' line form, when in fact it is the general principle that produces the correct condition.

Interesting Facts

The condition $pp' + rr' = -1$ for perpendicular lines in this symmetric form is a standard result covered in the NCERT Class 12 Three Dimensional Geometry chapter.

3D perpendicularity conditions derived from dot products of direction vectors are the same mathematical idea used in computer graphics to test whether two edges or normal vectors of a 3D model are at right angles.

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Frequently Asked Questions

What is the correct condition for the two lines in Q20 to be perpendicular?

$pp' + rr' = -1$ , obtained by setting the dot product of direction vectors $(p,1,r)$ and $(p',1,r')$ to zero: $pp' + 1 + rr' = 0$ .

Why is direction vector (p,1,r) used instead of (p,0,r)?

Because in the given line equations $x = py+q$ , $z = ry+s$ , the variable y itself is the natural parameter (coefficient 1), while x and z are expressed in terms of y — so the direction vector's y-component is 1, not 0.