Assertion(A): Given two non-zero vectors 𝑎⃗ and 𝑏⃗⃗ . If 𝑟⃗ is another non -zero vector such that 𝑟⃗ × (𝑎⃗ + 𝑏⃗⃗) = 0⃗⃗ . Then 𝑟⃗ is perpendicular to 𝑎⃗ × 𝑏⃗⃗ . Reason (R): The vector (𝑎⃗ + 𝑏⃗⃗) is perpendicular to the plane of 𝑎⃗ and 𝑏⃗⃗ 1

Assertion(A): Given two non-zero vectors 𝑎⃗ and 𝑏⃗⃗ . If 𝑟⃗ is another non -zero vector such that 𝑟⃗ × (𝑎

Question

Q. Assertion(A): Given two non-zero vectors 𝑎⃗ and 𝑏⃗⃗ . If 𝑟⃗ is another non -zero vector such that 𝑟⃗ \times (𝑎⃗ + 𝑏⃗⃗) = 0⃗⃗ . Then 𝑟⃗ is perpendicular to 𝑎⃗ \times 𝑏⃗⃗ . Reason (R): The vector (𝑎⃗ + 𝑏⃗⃗) is perpendicular to the plane of 𝑎⃗ and 𝑏⃗⃗ 1

CBSESample paperSubjective· 1mImportance★★★★★

✓ Free question

Concept understanding — Perpendicular Vectors Condition

Perpendicular Vectors Condition

Two arrows that meet at a right angle — one east, one north — are perpendicular (or orthogonal) vectors. How do you check this without a protractor, especially in 3D where the angle is hard to draw?

The Idea: Zero Overlap

When two vectors are perpendicular, neither "borrows" any length from the other: walking along one makes zero progress in the direction of the other. The tool that measures this overlap is the dot product.

\vec{a}\perp\vec{b} \iff \vec{a}\cdot\vec{b}=0

Why? Using $\vec{a}\cdot\vec{b}=\lVert\vec{a}\rVert\,\lVert\vec{b}\rVert\cos\theta$ , a right angle gives $\cos 90^\circ=0$ , so the dot product vanishes. In coordinates, for $\vec{a}=(a_1,a_2,a_3)$ and $\vec{b}=(b_1,b_2,b_3)$ ,

\vec{a}\cdot\vec{b}=a_1b_1+a_2b_2+a_3b_3,

and you simply check whether this sum is $0$ .

Examples

2D: $\vec{a}=(3,4)$ , $\vec{b}=(4,-3)$ : $\;3(4)+4(-3)=12-12=0$ — perpendicular. (In general $(x,y)$ and $(y,-x)$ are always perpendicular.)

3D: $\vec{p}=(1,2,3)$ , $\vec{q}=(2,-1,0)$ : $\;2-2+0=0$ — perpendicular.

Not every pair qualifies: $(2,1)\cdot(1,3)=2+3=5\neq 0$ , so those two are not perpendicular.

Watch out

In dimensions above 3 we cannot picture the right angle, but the test is unchanged: dot product $=0$ still defines orthogonality.

Why It Matters

The condition appears constantly — finding a line perpendicular to another, showing two lines or planes meet at right angles, and physics (a force perpendicular to displacement does zero work). Whenever you read "perpendicular" or "orthogonal," think dot product = 0.

Tip

To build a vector perpendicular to a given $\vec{a}$ , solve $\vec{a}\cdot\vec{x}=0$ — there are infinitely many solutions, all lying in the plane perpendicular to $\vec{a}$ .

The dot-product-equals-zero test for perpendicular vectors is one of the most heavily tested facts in the NCERT Class 12 Vector Algebra chapter, appearing across CBSE board papers, JEE Main and state CET vector questions. "Condition for two vectors to be perpendicular" is a top search term, and this single formula underlies work-done and right-angle proof questions throughout Class 12 Physics and Maths alike.

The key idea is that $\vec{r}$ is parallel to $\vec{a}+\vec{b}$ (from the cross product being zero), and $\vec{a}\times\vec{b}$ is perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ . Since $\vec{a}+\vec{b}$ lies in that same plane, $\vec{r}$ is perpendicular to $\vec{a}\times\vec{b}$ . Assertion is true, Reason is false.

Let’s unpack this carefully. The question tests two things: the geometric meaning of a cross product being zero, and the direction of the sum of two vectors relative to their cross product.

The core idea: If $\vec{r} \times (\vec{a}+\vec{b}) = \vec{0}$ , then $\vec{r}$ is parallel to $\vec{a}+\vec{b}$ (provided both are non-zero). Meanwhile, $\vec{a} \times \vec{b}$ is perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ . The sum $\vec{a}+\vec{b}$ lies in that same plane. So $\vec{r}$ , being parallel to the sum, is also in that plane — hence perpendicular to $\vec{a}\times\vec{b}$ .

Now, the Reason says: “The vector $(\vec{a}+\vec{b})$ is perpendicular to the plane of $\vec{a}$ and $\vec{b}$ .” That’s the opposite of the truth — the sum lies in the plane, not perpendicular to it. So the Reason is false.

Let’s go step by step.

What does $\vec{r} \times (\vec{a}+\vec{b}) = \vec{0}$ tell us? For two non-zero vectors, their cross product is zero if and only if they are parallel (or one is zero). Since $\vec{r}$ and $\vec{a}+\vec{b}$ are both given as non-zero, we get:

\vec{r} \parallel (\vec{a}+\vec{b})

That is, $\vec{r}$ is some scalar multiple of $\vec{a}+\vec{b}$ .

Where does $\vec{a}+\vec{b}$ lie?

The sum of two vectors always lies in the plane spanned by them. Think of it geometrically: if you place $\vec{a}$ and $\vec{b}$ tail-to-tail, their sum is the diagonal of the parallelogram they form — that diagonal is definitely in the same plane as the two original vectors. So $\vec{a}+\vec{b}$ is in the plane of $\vec{a}$ and $\vec{b}$ .
What about $\vec{a} \times \vec{b}$ ?

By definition, the cross product of two non-zero, non-parallel vectors is a vector perpendicular to the plane containing them. So $\vec{a} \times \vec{b}$ is perpendicular to the plane of $\vec{a}$ and $\vec{b}$ .
Putting it together:

Since $\vec{r}$ is parallel to $\vec{a}+\vec{b}$ , and $\vec{a}+\vec{b}$ lies in the plane of $\vec{a}$ and $\vec{b}$ , it follows that $\vec{r}$ also lies in that plane. A vector in the plane is perpendicular to the plane’s normal vector, which is $\vec{a} \times \vec{b}$ . Hence:

\vec{r} \perp (\vec{a} \times \vec{b})

The Assertion is true.

Checking the Reason: The Reason claims $\vec{a}+\vec{b}$ is perpendicular to the plane of $\vec{a}$ and $\vec{b}$ . That’s false — it’s actually in the plane. The only vector perpendicular to that plane (from the given set) is $\vec{a} \times \vec{b}$ (or any scalar multiple of it). So the Reason is false.

Watch out

A common mistake is to think that $\vec{a}+\vec{b}$ is perpendicular to the plane because it “looks like” a cross product result. But the sum is a linear combination, not a cross product — it stays in the plane.

Tip

A quick way to remember: The cross product of two vectors gives a vector out of their plane. Their sum stays in the plane. So any vector parallel to the sum is automatically perpendicular to the cross product.

✓Final answer

The Assertion is true but the Reason is false.

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

Showing the 12 most recent of 39 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$p$
›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.

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

Recall the dot product formula.

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

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

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

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

Simplify the equation.

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

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
$p = \frac{5}{2}$
CBSE 20261 mark
$x = py + q, z = ry + s$
›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
$pp' + rr' = -1$
CBSE 2026Set 65/1/11 markMCQ
$x = py + q$
›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 $\vec{b}_1 \cdot \vec{b}_2 = 0$ ."
$\vec{r} = \vec{a} + \lambda\vec{b}$
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.

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:
$pp' + rr' = -1$

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
$The correct option is (D) .$
CBSE 2026Set A1 markMCQ
$3\vec{i} + \vec{j} - 2\vec{k}$
(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
$-9$
CBSE 2026Set A1 markMCQ
$|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$
(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
$\vec{a}\perp\vec{b}$
CBSE 2026Set A1 markMCQ
$x - 4y + \lambda z + 3 = 0$
(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
$2$
CBSE 2026Set A1 markMCQ
$\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$
(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
$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$
CBSE 2026Set ANNUAL1 markMCQ
$\frac{x+1}{1}=\frac{y+2}{\lambda}=\frac{z-1}{-1}$
(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
$\lambda=1$
CBSE 2026Set ANNUAL1 markMCQ
$x$
(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
$-13$
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
$\lambda = 2/5$
CBSE 2026Set ANNUAL1 markMCQ
$x=ay+b,\ z=cy+d$
(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
$aa_{1}+cc_{1}=-1$
CBSE 2025Set E1 markMCQ
$5, 2, 4$
(a) $9$
(b) $-9$
(c) $8$
(d) $-8$
›Reveal solutionSolution

Perpendicular lines: $a_1a_2+b_1b_2+c_1c_2=0\Rightarrow 20+16+4x=0\Rightarrow x=-9$ .

Two lines are perpendicular when the sum of the products of corresponding direction ratios is zero:

$5(4) + 2(8) + 4(x) = 0$

$20 + 16 + 4x = 0$

$4x = -36 \Rightarrow x = -9.$

✓Final answer
$-9$

🎓Unlock everything free for 14 days

Unlock everything free for 14 days