Question:
Vector of magnitude 3 making equal angles with $x$ and $y$ axes and perpendicular to $z$ axis is
Vector of magnitude 3 making equal angles with $x$ and $y$ axes and perpendicular to $z$ axis is
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If a vector makes equal angles with two axes, the corresponding components are equal. If it is perpendicular to an axis, the component along that axis becomes zero.
Updated On: Mar 10, 2026
- $\hat{i}+2\sqrt{2}\hat{j}$
- $3\hat{k}$
- $\dfrac{3\sqrt{2}}{2}\hat{i}+\dfrac{3\sqrt{2}}{2}\hat{j}$
- $\sqrt{3}\hat{i}+\sqrt{3}\hat{j}+\sqrt{3}\hat{k}$
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The Correct Option is C
Solution and Explanation
Step 1: Understand the conditions of the vector.
The vector has magnitude \[ |\vec{v}|=3 \] It makes equal angles with the \(x\)-axis and \(y\)-axis.
Also it is perpendicular to the \(z\)-axis.
Step 2: Use perpendicular condition.
If a vector is perpendicular to the \(z\)-axis, its \(k\)-component must be zero.
Thus the vector is of the form \[ \vec{v}=a\hat{i}+a\hat{j}+0\hat{k} \] because it makes equal angles with \(x\) and \(y\) axes.
Step 3: Use magnitude formula.
\[ |\vec{v}|=\sqrt{a^2+a^2} \] \[ 3=\sqrt{2a^2} \] \[ 3=\sqrt{2}\,a \] \[ a=\frac{3}{\sqrt{2}} \] Step 4: Simplify the vector.
\[ \vec{v}=\frac{3}{\sqrt{2}}\hat{i}+\frac{3}{\sqrt{2}}\hat{j} \] Rationalizing denominator:
\[ \frac{3}{\sqrt{2}}=\frac{3\sqrt{2}}{2} \] Thus \[ \vec{v}=\frac{3\sqrt{2}}{2}\hat{i}+\frac{3\sqrt{2}}{2}\hat{j} \] Conclusion:
Hence the required vector is \[ \frac{3\sqrt{2}}{2}\hat{i}+\frac{3\sqrt{2}}{2}\hat{j} \] Final Answer: $\boxed{\dfrac{3\sqrt{2}}{2}\hat{i}+\dfrac{3\sqrt{2}}{2}\hat{j}}$
The vector has magnitude \[ |\vec{v}|=3 \] It makes equal angles with the \(x\)-axis and \(y\)-axis.
Also it is perpendicular to the \(z\)-axis.
Step 2: Use perpendicular condition.
If a vector is perpendicular to the \(z\)-axis, its \(k\)-component must be zero.
Thus the vector is of the form \[ \vec{v}=a\hat{i}+a\hat{j}+0\hat{k} \] because it makes equal angles with \(x\) and \(y\) axes.
Step 3: Use magnitude formula.
\[ |\vec{v}|=\sqrt{a^2+a^2} \] \[ 3=\sqrt{2a^2} \] \[ 3=\sqrt{2}\,a \] \[ a=\frac{3}{\sqrt{2}} \] Step 4: Simplify the vector.
\[ \vec{v}=\frac{3}{\sqrt{2}}\hat{i}+\frac{3}{\sqrt{2}}\hat{j} \] Rationalizing denominator:
\[ \frac{3}{\sqrt{2}}=\frac{3\sqrt{2}}{2} \] Thus \[ \vec{v}=\frac{3\sqrt{2}}{2}\hat{i}+\frac{3\sqrt{2}}{2}\hat{j} \] Conclusion:
Hence the required vector is \[ \frac{3\sqrt{2}}{2}\hat{i}+\frac{3\sqrt{2}}{2}\hat{j} \] Final Answer: $\boxed{\dfrac{3\sqrt{2}}{2}\hat{i}+\dfrac{3\sqrt{2}}{2}\hat{j}}$
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