Vector of magnitude 3 making equal angles with $x$ and $y$ axes and perpendicular to $z$ axis is (A) $\hat{i} + 2\sqrt{2}\, \hat{j}$ (B) $3\hat{k}$ (C) $\frac{3\sqrt{2}}{2} \hat{i} + \frac{3\sqrt{2}}{2} \hat{j}$ (D) $\sqrt{3}\, \hat{i} + \sqrt{3}\, \hat{j} + \sqrt{3}\, \hat{k}$

> [!TLDR] > The vector makes equal angles with the $x$ and $y$ axes and is perpendicular to the $z$ axis, so its $z$-component is zero and its $x$ and $y$ components are equal. With magnitude 3, each component is $\frac{3}{\sqrt{2}}$, giving the vector $\frac{3}{\sqrt{2}} \hat{i} + \frac{3}{\sqrt{2}} \hat{j}$, which matches option **(C)** after rationalising. The key idea here is that a vector’s direction cosines tell you how it’s oriented relative to the axes. If a vector makes equal angles with the $x$ and $y$ axes, its direction cosines (and hence its components) along those axes are equal. And if it’s perpendicular to the $z$ axis, its $z$-component is zero — the vector lies entirely in the $xy$-plane. So we’re looking for a vector of the form $a\hat{i} + a\hat{j} + 0\hat{k}$, where $a$ is the (equal) $x$ and $y$ components. The magnitude condition then fixes $a$. Let’s work through it. 1. **Set up the vector form.** Let the vector be $\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$. Perpendicular to the $z$ axis means $v_z = 0$. Equal angles with $x$ and $y$ axes means the direction cosines $\cos\alpha$ and $\cos\beta$ are equal. Since direction cosines are proportional to the components, we have $v_x = v_y$. So $\vec{v} = a\hat{i} + a\hat{j}$, where $a = v_x = v_y$. 2. **Apply the magnitude condition.** The magnitude is given as 3: $$|\vec{v}| = \sqrt{a^2 + a^2 + 0^2} = \sqrt{2a^2} = |a|\sqrt{2} = 3.$$ Since magnitude is positive, $a$ must be positive (the vector’s direction is fixed by the signs, but the problem doesn’t specify a sign, so we take the positive one). $$a\sqrt{2} = 3 \quad\Rightarrow\quad a = \frac{3}{\sqrt{2}}.$$ 3. **Write the vector.** $$\vec{v} = \frac{3}{\sqrt{2}} \hat{i} + \frac{3}{\sqrt{2}} \hat{j}.$$ This is not yet in the form of any option — but option (C) is $\frac{3\sqrt{2}}{2} \hat{i} + \frac{3\sqrt{2}}{2} \hat{j}$. Notice that $\frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$ (mul

Vector of magnitude 3 making equal angles with x and y axes and perpendicular to z axis is (A) hati + 2√(2) ha

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The vector makes equal angles with the $x$ and $y$ axes and is perpendicular to the $z$ axis, so its $z$ -component is zero and its $x$ and $y$ components are equal. With magnitude 3, each component is $\frac{3}{\sqrt{2}}$ , giving the vector $\frac{3}{\sqrt{2}} \hat{i} + \frac{3}{\sqrt{2}} \hat{j}$ , which matches option (C) after rationalising.

The key idea here is that a vector’s direction cosines tell you how it’s oriented relative to the axes. If a vector makes equal angles with the $x$ and $y$ axes, its direction cosines (and hence its components) along those axes are equal. And if it’s perpendicular to the $z$ axis, its $z$ -component is zero — the vector lies entirely in the $xy$ -plane.

So we’re looking for a vector of the form $a\hat{i} + a\hat{j} + 0\hat{k}$ , where $a$ is the (equal) $x$ and $y$ components. The magnitude condition then fixes $a$ .

Let’s work through it.

Set up the vector form.

Let the vector be $\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$ .

Perpendicular to the $z$ axis means $v_z = 0$ .

Equal angles with $x$ and $y$ axes means the direction cosines $\cos\alpha$ and $\cos\beta$ are equal. Since direction cosines are proportional to the components, we have $v_x = v_y$ .

So $\vec{v} = a\hat{i} + a\hat{j}$ , where $a = v_x = v_y$ .
Apply the magnitude condition.

The magnitude is given as 3:

|\vec{v}| = \sqrt{a^2 + a^2 + 0^2} = \sqrt{2a^2} = |a|\sqrt{2} = 3.

Since magnitude is positive, $a$ must be positive (the vector’s direction is fixed by the signs, but the problem doesn’t specify a sign, so we take the positive one).

a\sqrt{2} = 3 \quad\Rightarrow\quad a = \frac{3}{\sqrt{2}}.

Write the vector.

\vec{v} = \frac{3}{\sqrt{2}} \hat{i} + \frac{3}{\sqrt{2}} \hat{j}.

This is not yet in the form of any option — but option (C) is $\frac{3\sqrt{2}}{2} \hat{i} + \frac{3\sqrt{2}}{2} \hat{j}$ . Notice that $\frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$ (multiply numerator and denominator by $\sqrt{2}$ ). So they are identical. …

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

CBSE 2026Set 65/2/11 markMCQ
$x$
›Reveal solutionSolution
The vector makes equal angles with the $x$ and $y$ axes and is perpendicular to the $z$ axis, so its $z$ -component is zero and its $x$ and $y$ components are equal. With magnitude 3, each component is $\frac{3}{\sqrt{2}}$ , giving the vector $\frac{3}{\sqrt{2}} \hat{i} + \frac{3}{\sqrt{2}} \hat{j}$ , which matches option (C) after rationalising.

The key idea here is that a vector’s direction cosines tell you how it’s oriented relative to the axes. If a vector makes equal angles with the $x$ and $y$ axes, its direction cosines (and hence its components) along those axes are equal. And if it’s perpendicular to the $z$ axis, its $z$ -component is zero — the vector lies entirely in the $xy$ -plane.

So we’re looking for a vector of the form $a\hat{i} + a\hat{j} + 0\hat{k}$ , where $a$ is the (equal) $x$ and $y$ components. The magnitude condition then fixes $a$ .

Let’s work through it.

Set up the vector form.

Let the vector be $\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$ .

Perpendicular to the $z$ axis means $v_z = 0$ .

Equal angles with $x$ and $y$ axes means the direction cosines $\cos\alpha$ and $\cos\beta$ are equal. Since direction cosines are proportional to the components, we have $v_x = v_y$ .

So $\vec{v} = a\hat{i} + a\hat{j}$ , where $a = v_x = v_y$ .

Apply the magnitude condition.

The magnitude is given as 3:

$|\vec{v}| = \sqrt{a^2 + a^2 + 0^2} = \sqrt{2a^2} = |a|\sqrt{2} = 3.$
Since magnitude is positive, $a$ must be positive (the vector’s direction is fixed by the signs, but the problem doesn’t specify a sign, so we take the positive one).
$a\sqrt{2} = 3 \quad\Rightarrow\quad a = \frac{3}{\sqrt{2}}.$

Write the vector.

$\vec{v} = \frac{3}{\sqrt{2}} \hat{i} + \frac{3}{\sqrt{2}} \hat{j}.$
This is not yet in the form of any option — but option (C) is $\frac{3\sqrt{2}}{2} \hat{i} + \frac{3\sqrt{2}}{2} \hat{j}$ . Notice that $\frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$ (multiply numerator and denominator by $\sqrt{2}$ ). So they are identical. …
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CBSE 2026Set V11 markMCQ
$x$
(a) $0, k, 0$
(b) $0, 0, k$
(c) $k, 0, 0$
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The $x$ -axis points along $\hat i=(1,0,0)$ , so its direction ratios are $k,0,0$ ; answer (c).

The positive $x$ -axis is directed along the unit vector $\hat i=(1,0,0)$ . Direction ratios are any nonzero scalar multiple of the d …
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(d) <9, -2, 5>
›Reveal solutionSolution

In the vector form $\vec r = \vec a + m\vec b$ , the coefficients of the parameter $m$ (i.e. $\vec b$ ) directly give the direction ratios of the line.

The line is $\vec r = 2\hat i - 3\hat j + \hat k + m(9\hat i - 2\hat j + 5\hat k)$ .

…

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CBSE 2025Set ANNUAL1 mark
$Q. Direction ratios of straight line r⃗ = 4î - 7ĵ + 5k̂ + s(9î - 2ĵ + 5k̂) are ______.$
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In the vector form $\vec r = \vec a + s\,\vec b$ , the coefficients of the parameter give the line's direction ratios.

For $\vec r = 4\hat i - 7\hat j + 5\hat k + s(9\hat i - 2\hat j + 5\hat k)$ , the fixed point is $(4,-7,5)$ and the direction vector is …

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CBSE 2024Set ANNUAL1 mark
$(2, 3, -4)$
›Reveal solutionSolution

Direction ratios of a line joining $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are $(x_2-x_1,\,y_2-y_1,\,z_2-z_1)$ .

Let the points be $P(2,3,-4)$ and $Q(1,3,-2)$ .

Direction ratios of line PQ: …

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CBSE 2022Set ANNUAL1 markMCQ
$3x + 4y + 5z - 6 = 0$
(a) $3, 4, 5$
(b) $-3, 4, 5$
(c) $3, -4, 5$
(d) $2, 3, -4$
›Reveal solutionSolution

For $ax+by+cz+d=0$ , the normal has direction ratios $(a,b,c)=(3,4,5)$ .

The normal to a plane $ax+by+cz+d=0$ has direction ratios equal to the coefficients $(a,b,c)$ …

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CBSE 2019Set HE1 mark
$Q. Fill in the blank: Direction ratio of two parallel lines will be ______.$
›Reveal solutionSolution

Parallel lines have the same direction, so their direction ratios are proportional.

Direction ratios $(a,b,c)$ describe the direction a line points in. Two lines are parallel precisely when they point in the same (or exactly opposite) direction, which means one line's direction ratios are a scalar multiple of the oth …

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CBSE 2018Set ANNUAL1 markMCQ
$Q. Direction ratios of the line joining points A(2, 3, -4) and B(1, -2, 3) is:$
(a)
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(b)
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(c)
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(d)
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Direction ratios $=(-1,-5,7)$ — option (iv).

Concept. For a line joining $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ , the direction ratios are $x_2-x_1,\ y_2-y_1,\ z_2-z_1$ .

Steps. With $A(2,3,-4)$ and $B(1,-2,3)$ : …

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Q.Vector of magnitude 3 making equal angles with $x$ and $y$ axes and perpendicular to $z$ axis is (A) $\hat{i} + 2\sqrt{2}\, \hat{j}$ (B) $3\hat{k}$ (C) $\frac{3\sqrt{2}}{2} \hat{i} + \frac{3\sqrt{2}}{2} \hat{j}$ (D) $\sqrt{3}\, \hat{i} + \sqrt{3}\, \hat{j} + \sqrt{3}\, \hat{k}$

Concept understanding — Direction Vectors

Direction Vectors

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Vector Equation of a Line

Direction Ratios and Direction Cosines

Why It Matters …

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