Question:
If the vector \(\vec b=3\hat j+4\hat k\) is written as the sum of a vector \(\vec b_1\), parallel to \(\vec a=\hat i+\hat j\) and a vector \(\vec b_2\), perpendicular to \(\vec a\), then \(\vec b_1\times\vec b_2\) is equal to :
If the vector \(\vec b=3\hat j+4\hat k\) is written as the sum of a vector \(\vec b_1\), parallel to \(\vec a=\hat i+\hat j\) and a vector \(\vec b_2\), perpendicular to \(\vec a\), then \(\vec b_1\times\vec b_2\) is equal to :
Show Hint
Let \(\vec b_1=t(\hat i+\hat j)\). Then \(\vec b_2=\vec b-\vec b_1=-t\hat i+(3-t)\hat j+4\hat k\).
\(\vec b_2\cdot\vec a=0\Rightarrow -t+3-t=0\Rightarrow t=\frac32\).
\(\vec b_1=\frac32\hat i+\frac32\hat j\), \(\vec b_2=-\frac32\hat i+\frac32\hat j+4\hat k\).
\(\vec b_1\times\vec b_2\): \(\hat i\)-component \(=\frac32\cdot4-0=6\); \(\hat j\)-component \(=-\left(\frac32\cdot4-0\right)=-6\); \(\hat k\)-component \(=\frac32\cdot\frac32-\frac32\cdot\left(-\frac32\right)=\frac92\).
So \(\vec b_1\times\vec b_2=6\hat i-6\hat j+\frac92\hat k\).
Answer: (2) \(6\hat i-6\hat j+\frac92\hat k\)
\(\vec b_2\cdot\vec a=0\Rightarrow -t+3-t=0\Rightarrow t=\frac32\).
\(\vec b_1=\frac32\hat i+\frac32\hat j\), \(\vec b_2=-\frac32\hat i+\frac32\hat j+4\hat k\).
\(\vec b_1\times\vec b_2\): \(\hat i\)-component \(=\frac32\cdot4-0=6\); \(\hat j\)-component \(=-\left(\frac32\cdot4-0\right)=-6\); \(\hat k\)-component \(=\frac32\cdot\frac32-\frac32\cdot\left(-\frac32\right)=\frac92\).
So \(\vec b_1\times\vec b_2=6\hat i-6\hat j+\frac92\hat k\).
Answer: (2) \(6\hat i-6\hat j+\frac92\hat k\)
Updated On: Nov 11, 2026
- \(-3\hat i+3\hat j-9\hat k\)
- \(6\hat i-6\hat j+\dfrac92\hat k\)
- \(-6\hat i+6\hat j-\dfrac92\hat k\)
- \(3\hat i-3\hat j+9\hat k\)
Show Solution
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The Correct Option is B
Solution and Explanation
Take \(\vec b_1=t\vec a\), \(\vec b_2=\vec b-\vec b_1\) and impose \(\vec b_2\cdot\vec a=0\).
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