Q.Show that the line through the points $(1, -1, 2), (3, 4, -2)$ is perpendicular to the line through the points $(0, 3, 2)$ and $(3, 5, 6)$.

Showing the 12 most recent of 39 on this concept.

1. CBSE 20261 mark
   Q.Assertion (A): Lines given by $x = py + q, z = ry + s$ and $x = p'y + q', z = r'y + s'$ are perpendicular if $pp' + rr' = 1$. 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 if $\vec{b}_1 \cdot \vec{b}_2 = 0$.

   ›Reveal solutionSolution

   The key idea is to convert the given symmetric equations into vector form, extract the direction vectors, and apply the perpendicularity condition $\vec{b}_1 \cdot \vec{b}_2 = 0$. The assertion is false because the correct condition is $pp' + rr' = -1$, not $+1$.

   Let’s understand why. The problem tests two things: first, how to read direction vectors from a pair of linear equations representing a line, and second, the precise condition for perpendicular lines in 3D.

   The Core Concept

   Two lines in space are perpendicular when their direction vectors are orthogonal — that is, their dot product is zero. The Reason (R) states this correctly: for lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$, perpendicularity means $\vec{b}_1 \cdot \vec{b}_2 = 0$.

   The trick lies in the Assertion (A). The given equations $x = py + q, z = ry + s$ represent a line, but not in the standard symmetric form. We need to extract its direction vector.

   Watch out

   A common mistake is to read the coefficients of $y$ directly as direction ratios. That would give $(p, 1, r)$, but this is incorrect — the equations are not in the form $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$.

   Let’s work through the extraction properly.

   Step-by-Step Solution

   1. Rewrite the line in symmetric form

   The equations $x = py + q$ and $z = ry + s$ both express $x$ and $z$ in terms of $y$. This means $y$ acts as a parameter. Let $y = t$. Then:

   • $x = p t + q$
   • $y = t$
   • $z = r t + s$

   So the parametric form is:

   $(x, y, z) = (q, 0, s) + t(p, 1, r)$

   The direction vector of the first line is $\vec{b}_1 = (p, 1, r)$.

   2. Similarly for the second line

   For $x = p'y + q', z = r'y + s'$, let $y = u$. Then:

   • $x = p' u + q'$
   • $y = u$
   • $z = r' u + s'$

   So the direction vector is $\vec{b}_2 = (p', 1, r')$.

   3. Apply the perpendicular condition

   For perpendicular lines, $\vec{b}_1 \cdot \vec{b}_2 = 0$:

   $(p, 1, r) \cdot (p', 1, r') = p p' + (1)(1) + r r' = p p' + r r' + 1 = 0$

   Therefore:

   $p p' + r r' = -1$

   Tip

   Notice the $+1$ comes from the product of the $y$-coefficients (both are $1$). This is the hidden term that changes the condition from $+1$ to $-1$.

   4. Compare with the assertion

   The Assertion (A) claims the condition is $pp' + rr' = 1$. We have derived $pp' + rr' = -1$. So (A) is false.

   The Reason (R) is a standard, correct statement about perpendicular lines in vector form. So (R) is true.

   5. Final classification

   Since (A) is false and (R) is true, the correct choice is: Assertion (A) is false, but Reason (R) is true.

   ✓Final answer

   The Assertion is false and the Reason is true; the correct condition is $pp' + rr' = -1$, not $+1$.

2. CBSE 2026Set 65/1/11 markMCQ
   Q.Assertion (A): The lines $x = py + q$, $z = ry + s$ 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 $\vec{b}_1 \cdot \vec{b}_2 = 0$. (A) Both Assertion (A) and Reason (R) are true and 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.

   ›Reveal solutionSolution

   The condition for two lines to be perpendicular is that the dot product of their direction vectors is zero. Reason (R) correctly states this. Assertion (A) provides an incorrect condition for perpendicularity based on the given line equations. Therefore, Assertion (A) is false, and Reason (R) is true.

   The core concept for determining if two lines are perpendicular in 3D space relies on their direction vectors. A line's direction vector indicates the path it follows. If two lines are perpendicular, their direction vectors must also be perpendicular. The mathematical condition for two non-zero vectors to be perpendicular is that their dot product is zero. This is a fundamental property of the dot product.

   Let's evaluate the given Assertion and Reason.

   Evaluating Reason (R)

   Reason (R) states: "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 vector equation of a line is $\vec{r} = \vec{a} + \lambda\vec{b}$, where $\vec{a}$ is the position vector of a point on the line and $\vec{b}$ is the direction vector of the line.

   The vectors $\vec{b}_1$ and $\vec{b}_2$ are the direction vectors of the respective lines. If the lines are perpendicular, their direction vectors must be perpendicular. The dot product of two perpendicular vectors is indeed zero. This statement is a correct and fundamental principle in vector algebra and 3D geometry.

   Therefore, Reason (R) is true.

   Evaluating Assertion (A)

   Assertion (A) states: "The lines $x = py + q$, $z = ry + s$ and $x = p'y + q'$, $z = r'y + s'$ are perpendicular to each other when $pp' + rr' = 1$."

   To check this assertion, we first need to find the direction vectors for each line from their given equations. The equations are given in a non-standard form, so we will convert them to the symmetric form $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$, where $\langle a, b, c \rangle$ is the direction vector.

   1. Find the direction vector for the first line: The equations are $x = py + q$ and $z = ry + s$. From $x = py + q$, we can write $y = \frac{x-q}{p}$. From $z = ry + s$, we can write $y = \frac{z-s}{r}$. Combining these, we get the symmetric form:

   $\frac{x-q}{p} = \frac{y-0}{1} = \frac{z-s}{r}$

   The direction vector for the first line, $\vec{b}_1$, is $\langle p, 1, r \rangle$.
   

   2. Find the direction vector for the second line:

   The equations are $x = p'y + q'$ and $z = r'y + s'$.

   Similarly, from $x = p'y + q'$, we get $y = \frac{x-q'}{p'}$.

   From $z = r'y + s'$, we get $y = \frac{z-s'}{r'}$.

   Combining these, we get the symmetric form:

   $\frac{x-q'}{p'} = \frac{y-0}{1} = \frac{z-s'}{r'}$

   The direction vector for the second line, $\vec{b}_2$, is $\langle p', 1, r' \rangle$.
   

   3. Apply the perpendicularity condition:

   According to Reason (R), for the two lines to be perpendicular, the dot product of their direction vectors must be zero: $\vec{b}_1 \cdot \vec{b}_2 = 0$.

   Let's calculate the dot product:

   $\vec{b}_1 \cdot \vec{b}_2 = (p)(p') + (1)(1) + (r)(r')$

   $\vec{b}_1 \cdot \vec{b}_2 = pp' + 1 + rr'$

   For the lines to be perpendicular, we must have:
   

   $pp' + 1 + rr' = 0$

   This simplifies to:
   

   $pp' + rr' = -1$

   4. Compare with the assertion: Assertion (A) states that the lines are perpendicular when $pp' + rr' = 1$. Our derivation shows that the correct condition for perpendicularity is $pp' + rr' = -1$. Since the condition given in Assertion (A) ($pp' + rr' = 1$) is different from the correct condition ($pp' + rr' = -1$), Assertion (A) is false.

   Watch out

   A common mistake is to directly assume the coefficients $p, r$ and $p', r'$ are the components of the direction vectors. Remember that the standard symmetric form is $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$, where the denominators $a, b, c$ form the direction vector. In this problem, $y$ is the common parameter, leading to the $y$-component of the direction vector being $1$.

   Conclusion

   Reason (R) is true, as it states a fundamental condition for perpendicular lines.

   Assertion (A) is false, because the derived condition for perpendicularity is $pp' + rr' = -1$, not $pp' + rr' = 1$.

   Therefore, Assertion (A) is false and Reason (R) is true.

   ✓Final answer

   The correct option is (D).

3. CBSE 2026Set 65/1/11 markMCQ
   Q.The value of $p$ for which vectors $\hat{i} + 2\hat{j} + 3\hat{k}$ and $2\hat{i} - p\hat{j} + \hat{k}$ are perpendicular to each other is (A) $0$ (B) $1$ (C) $\dfrac{5}{2}$ (D) $-\dfrac{5}{2}$

   ›Reveal solutionSolution

   Two vectors are perpendicular when their dot product equals zero. Setting the dot product of $\hat{i} + 2\hat{j} + 3\hat{k}$ and $2\hat{i} - p\hat{j} + \hat{k}$ to zero gives $p = \frac{5}{2}$, which corresponds to option (C).

   The key idea here is the perpendicular vectors condition: two vectors are perpendicular (orthogonal) if and only if their dot product is zero. This is a fundamental geometric fact — the dot product measures how much one vector "projects" onto another; when they're at right angles, that projection is zero.

   Let’s apply this step by step.

   1. Write the vectors in component form.

      Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - p\hat{j} + \hat{k}$.

      In component notation:

      $\vec{a} = (1, 2, 3)$ and $\vec{b} = (2, -p, 1)$.

   2. Recall the dot product formula.

      For vectors $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$,

   $\vec{a} \cdot \vec{b} = x_1 x_2 + y_1 y_2 + z_1 z_2.$

   3. Set up the perpendicular condition. We require $\vec{a} \cdot \vec{b} = 0$. So:

   $(1)(2) + (2)(-p) + (3)(1) = 0.$

   4. Simplify the equation.

   $2 - 2p + 3 = 0 \quad \Rightarrow \quad 5 - 2p = 0.$

   5. Solve for $p$.

   $2p = 5 \quad \Rightarrow \quad p = \frac{5}{2}.$

   Watch out

   A common mistake is to forget the sign of the middle term: $2 \times (-p) = -2p$, not $+2p$. Double-check each term's sign.

   Tip

   You can verify your answer quickly: plug $p = \frac{5}{2}$ back into the dot product — you should get exactly zero. If you get anything else, recheck the arithmetic.

   ✓Final answer

   The value is $p = \frac{5}{2}$, which matches option (C).

4. CBSE 2023Set 65/1/11 markMCQ
   Q.The value of p for which the vectors $2\hat{i} + p\hat{j} + \hat{k}$ and $-4\hat{i} - 6\hat{j} + 26\hat{k}$ are perpendicular to each other, is : (A) 3 (B) -3 (C) $-\frac{17}{3}$ (D) $\frac{17}{3}$

   ›Reveal solutionSolution

   Two vectors are perpendicular when their dot product equals zero. Setting the dot product of the given vectors to zero and solving for $p$ gives $p = 3$, which corresponds to option (A).

   Concept and Intuition

   The condition for two vectors to be perpendicular (orthogonal) is one of the most fundamental ideas in vector algebra. When two vectors are perpendicular, the angle between them is $90^\circ$, and the cosine of $90^\circ$ is zero. Since the dot product of two vectors is defined as the product of their magnitudes times the cosine of the angle between them, a zero dot product directly signals perpendicularity.

   For vectors $\vec{a}$ and $\vec{b}$:

   $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$

   When $\theta = 90^\circ$, $\cos 90^\circ = 0$, so $\vec{a} \cdot \vec{b} = 0$.

   This is a clean, algebraic condition — no need to compute magnitudes or angles. You simply multiply corresponding components, add them, and set the sum to zero.

   Watch out

   A common mistake is to forget that the dot product involves all three components. Students sometimes multiply only the $\hat{i}$ and $\hat{j}$ components, leaving out the $\hat{k}$ term. Always check that you've included every component.

   Step-by-Step Solution

   1. Write the vectors in component form.

   Let $\vec{a} = 2\hat{i} + p\hat{j} + \hat{k}$ and $\vec{b} = -4\hat{i} - 6\hat{j} + 26\hat{k}$.

   In component notation:

   • $\vec{a} = (2, \; p, \; 1)$
   • $\vec{b} = (-4, \; -6, \; 26)$

   2. Apply the perpendicular condition.

   For perpendicular vectors: $\vec{a} \cdot \vec{b} = 0$.

   The dot product is computed component-wise:

   $\vec{a} \cdot \vec{b} = (2)(-4) + (p)(-6) + (1)(26)$

   3. Simplify the expression.

   $= -8 - 6p + 26$

   $= 18 - 6p$

   4. Set equal to zero and solve for $p$.

   $18 - 6p = 0$

   $6p = 18$

   $p = 3$

   Tip

   You can verify your answer quickly: if $p = 3$, then $\vec{a} = (2, 3, 1)$ and $\vec{b} = (-4, -6, 26)$. The dot product becomes $-8 - 18 + 26 = 0$, confirming perpendicularity. This check takes seconds and catches arithmetic errors.

   5. Match with the given options.

   The value $p = 3$ corresponds to option (A).

   ✓Final answer

   The value of $p$ is $\boxed{3}$, which is option (A).

5. CBSE 2026Set A1 markMCQ
   Q.If $3\vec{i} + \vec{j} - 2\vec{k}$ and $\vec{i} + \lambda\vec{j} - 3\vec{k}$ are perpendicular to each other then the value of $\lambda =$

   (a) $-3$

   (b) $-6$

   (c) $-9$

   (d) $-1$

   ›Reveal solutionSolution

   Set the dot product to zero.

   Perpendicular $\Rightarrow$ dot product $= 0$:

   $(3\vec{i}+\vec{j}-2\vec{k})\cdot(\vec{i}+\lambda\vec{j}-3\vec{k}) = 3(1) + 1(\lambda) + (-2)(-3) = 3 + \lambda + 6 = 9 + \lambda.$

   Setting $9 + \lambda = 0$ gives $\lambda = -9$.

   ✓Final answer

   (c) $-9$.

6. CBSE 2026Set A1 markMCQ
   Q.If $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$ then

   (a) $|\vec{a}| = |\vec{b}|$

   (b) $\vec{a} \parallel \vec{b}$

   (c) $\vec{a} \perp \vec{b}$

   (d) none of these

   ›Reveal solutionSolution

   $|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|$ means the vectors are perpendicular.

   Square both sides:

   $|\vec{a}+\vec{b}|^2 = |\vec{a}-\vec{b}|^2$

   $|\vec{a}|^2 + 2\vec{a}\cdot\vec{b} + |\vec{b}|^2 = |\vec{a}|^2 - 2\vec{a}\cdot\vec{b} + |\vec{b}|^2.$

   This gives $4\vec{a}\cdot\vec{b} = 0$, i.e. $\vec{a}\cdot\vec{b} = 0$, so $\vec{a} \perp \vec{b}$ (geometrically, the diagonals of the parallelogram are equal only for a rectangle).

   ✓Final answer

   (c) $\vec{a}\perp\vec{b}$.

7. CBSE 2026Set A1 markMCQ
   Q.If two planes $x - 4y + \lambda z + 3 = 0$ and $2x + 2y + 3z = 5$ are perpendicular to each other then $\lambda =$

   (a) $1$

   (b) $2$

   (c) $3$

   (d) $4$

   ›Reveal solutionSolution

   Normals must be perpendicular: their dot product is zero.

   Normals are $(1,-4,\lambda)$ and $(2,2,3)$. For the planes to be perpendicular:

   $1(2) + (-4)(2) + \lambda(3) = 0 \Rightarrow 2 - 8 + 3\lambda = 0 \Rightarrow 3\lambda = 6 \Rightarrow \lambda = 2.$

   ✓Final answer

   (b) $2$.

8. CBSE 2026Set A1 markMCQ
   Q.If the line $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ is parallel to the plane $a_2 x + b_2 y + c_2 z + d = 0$ then

   (a) $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

   (b) $a_1 x + b_1 y + c_1 z + d = 0$

   (c) $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$

   (d) none of these

   ›Reveal solutionSolution

   Line $\parallel$ plane $\Rightarrow$ line direction $\perp$ plane normal.

   The line has direction ratios $(a_1,b_1,c_1)$ and the plane has normal $(a_2,b_2,c_2)$. If the line is parallel to the plane, its direction lies in the plane, hence is perpendicular to the normal. So their dot product vanishes:

   $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0.$

   ✓Final answer

   (c) $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.

9. CBSE 2026Set ANNUAL1 markMCQ
   Q.If the straight lines $\frac{x+1}{1}=\frac{y+2}{\lambda}=\frac{z-1}{-1}$ and $\frac{x-1}{-\lambda}=\frac{y+1}{2}=\frac{z+1}{1}$ are perpendicular to each other, then the value of $\lambda$ is

   (a) $0$

   (b) $1$

   (c) $2$

   (d) $3$

   ›Reveal solutionSolution

   Two lines are perpendicular when the dot product of their direction ratios is zero.

   Direction ratios: line 1 = $(1,\lambda,-1)$, line 2 = $(-\lambda,2,1)$.

   Perpendicularity: $1(-\lambda)+\lambda(2)+(-1)(1)=0 \Rightarrow -\lambda+2\lambda-1=0 \Rightarrow \lambda-1=0 \Rightarrow \lambda=1$.

   ✓Final answer

   The correct option is (b) $\lambda=1$.

10. CBSE 2026Set ANNUAL1 markMCQ
    Q.For what value of $x$, vectors $x\hat{i}-3\hat{j}-5\hat{k}$ and $-\hat{i}+\hat{j}+2\hat{k}$ are perpendicular to each other?

    (a) 4

    (b) 7

    (c) -13

    (d) None of these

    ›Reveal solutionSolution

    Two vectors are perpendicular exactly when their dot product is zero.

    Let $\vec u = x\hat i - 3\hat j - 5\hat k$ and $\vec v = -\hat i+\hat j+2\hat k$.

    $\vec u \cdot \vec v = x(-1) + (-3)(1) + (-5)(2) = -x - 3 - 10 = -x-13$

    Setting this to $0$: $-x-13=0 \Rightarrow x=-13$.

    ✓Final answer

    (c) $-13$.

11. CBSE 2026Set ANNUAL1 markMCQ
    Q.If a⃗ and b⃗ are two vectors such that |a⃗| = 2 and |b⃗| = 5, then the value of λ for which a⃗ + λb⃗ and a⃗ − λb⃗ will be perpendicular is ................. .

    (a) 1/2

    (b) 2/3

    (c) 2/5

    (d) 3/4

    ›Reveal solutionSolution

    Perpendicularity means the dot product of $\vec a+\lambda\vec b$ and $\vec a-\lambda\vec b$ is zero.

    $(\vec a+\lambda\vec b)\cdot(\vec a-\lambda\vec b) = |\vec a|^2 - \lambda^2|\vec b|^2 = 0$

    $4 - 25\lambda^2 = 0 \Rightarrow \lambda^2 = \dfrac{4}{25} \Rightarrow \lambda = \dfrac{2}{5}$

    ✓Final answer

    $\lambda = 2/5$ — option (c).

12. CBSE 2026Set ANNUAL1 markMCQ
    Q.The two lines $x=ay+b,\ z=cy+d$ and $x=a_{1}y+b_{1},\ z=c_{1}y+d_{1}$ are perpendicular to each other, if

    (a) $\dfrac{a}{a_{1}}+\dfrac{c}{c_{1}}=1$

    (b) $\dfrac{a}{a_{1}}+\dfrac{c}{c_{1}}=-1$

    (c) $aa_{1}+cc_{1}=1$

    (d) $aa_{1}+cc_{1}=-1$

    ›Reveal solutionSolution

    Direction ratios are $(a,1,c)$ and $(a_1,1,c_1)$; perpendicularity gives $aa_1+1+cc_1=0$, i.e. $aa_1+cc_1=-1$.

    For the first line, taking $y=t$: $x=at+b,\ y=t,\ z=ct+d$, so its direction ratios are $(a,1,c)$.

    Similarly the second line has direction ratios $(a_{1},1,c_{1})$.

    The lines are perpendicular when the dot product of the direction vectors is zero:

    $aa_{1}+(1)(1)+cc_{1}=0\Rightarrow aa_{1}+cc_{1}=-1.$

    ✓Final answer

    $aa_{1}+cc_{1}=-1$ — option (D).