Question

Let \(\vec a,\vec b,\vec c\) be unit vectors such that \(\vec a\) is perpendicular to the plane containing \(\vec b\) and \(\vec c\), and angle between \(\vec b\) and \(\vec c\) is \(\frac{\pi}{3}\). Then \[ |\vec a+\vec b+\vec c|= \]

• A. \(3\)
• B. \(1\)
• C. \(2\)
• D. \(4\)

Hint:

If a vector is perpendicular to a plane, then it is perpendicular to every vector lying in that plane. Use dot product expansion to find magnitudes of sums of vectors.

Answer 1

The Correct Option is C

Step 1: Use the magnitude formula.
We need to find \[ |\vec a+\vec b+\vec c| \] Squaring both sides, \[ |\vec a+\vec b+\vec c|^2 = (\vec a+\vec b+\vec c)\cdot(\vec a+\vec b+\vec c) \] \[ = |\vec a|^2+|\vec b|^2+|\vec c|^2 +2\vec a\cdot\vec b +2\vec b\cdot\vec c +2\vec c\cdot\vec a \]

Step 2: Use the unit vector condition.
Since \(\vec a,\vec b,\vec c\) are unit vectors, \[ |\vec a|=|\vec b|=|\vec c|=1 \] Therefore, \[ |\vec a|^2=|\vec b|^2=|\vec c|^2=1 \]

Step 3: Use perpendicular condition.
Since \(\vec a\) is perpendicular to the plane containing \(\vec b\) and \(\vec c\), it is perpendicular to both \(\vec b\) and \(\vec c\).
Thus, \[ \vec a\cdot\vec b=0 \] and \[ \vec a\cdot\vec c=0 \]

Step 4: Use angle between \(\vec b\) and \(\vec c\).
Given angle between \(\vec b\) and \(\vec c\) is \[ \frac{\pi}{3} \] So, \[ \vec b\cdot\vec c=|\vec b||\vec c|\cos\frac{\pi}{3} \] \[ =1\cdot 1\cdot \frac12 \] \[ =\frac12 \]

Step 5: Substitute all values.
\[ |\vec a+\vec b+\vec c|^2 = 1+1+1+2(0)+2\left(\frac12\right)+2(0) \] \[ =3+1 \] \[ =4 \] Therefore, \[ |\vec a+\vec b+\vec c|=\sqrt4=2 \]

Step 6: Final conclusion.
Hence, \[ \boxed{2} \]