Which of the following statement are true?
A. The angle between the vectors \( 2\hat{\imath}+\hat{\jmath}+3\hat{k} \) and \( 3\hat{\imath}-2\hat{k} \) is \( \displaystyle\frac{\pi}{2} \)
B. If \( \vec{a} \) and \( \vec{b} \) are two non-zero vectors, then projection of \( \vec{b} \) on \( \vec{a} \) is \( \displaystyle\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|} \)
C. The unit vector normal to both vectors \( \vec{a}=\hat{\imath}-\hat{\jmath}-\hat{k} \) and \( \vec{b}=\hat{\imath}+\hat{\jmath}+\hat{k} \) is \( \displaystyle\frac{1}{\sqrt{2}}(-\hat{\jmath}+\hat{k}) \)
D. The area of triangle formed by adjacent sides represented by vectors \( \vec{a}=3\hat{\imath}+4\hat{\jmath} \) and \( \vec{b}=-5\hat{\imath}+7\hat{\jmath} \) is 41 sq. units
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A, B and C only
For (A), the dot product of the two vectors is
\( \displaystyle (2)(3)+(1)(0)+(3)(-2)=6+0-6=0 \)
For (C), a vector normal to both is the cross product:
\( \displaystyle \vec{a}\times\vec{b}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\ 1 & -1 & -1\\ 1 & 1 & 1\end{vmatrix} \)
\( \displaystyle =(-1+1)\hat{\imath}-(1+1)\hat{\jmath}+(1+1)\hat{k} \)
\( \displaystyle =-2\hat{\jmath}+2\hat{k} \)
\( \displaystyle \frac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}=\frac{-2\hat{\jmath}+2\hat{k}}{2\sqrt{2}} \)
\( \displaystyle =\frac{1}{\sqrt{2}}(-\hat{\jmath}+\hat{k}) \)
For (D), the area of a triangle on adjacent sides \( \vec{a},\vec{b} \) is \( \frac{1}{2}|\vec{a}\times\vec{b}| \):
\( \displaystyle \vec{a}\times\vec{b}=(3\cdot 7-4\cdot(-5))\hat{k}=41\hat{k} \)
\( \displaystyle \text{Area of triangle}=\frac{1}{2}|\vec{a}\times\vec{b}| \)
\( \displaystyle =\frac{41}{2}\text{ sq. units} \)
So the dot product in (A) is 0, the unit normal in (C) is \( \frac{1}{\sqrt{2}}(-\hat{\jmath}+\hat{k}) \), and the triangle in (D) has area \( \frac{41}{2} \) sq. units.
Checking the statements
(A) The angle between the vectors \( 2\hat{\imath}+\hat{\jmath}+3\hat{k} \) and \( 3\hat{\imath}-2\hat{k} \) is \( \frac{\pi}{2} \) — Correct. The dot product is 0, so the vectors are perpendicular.
(B) If \( \vec{a} \) and \( \vec{b} \) are two non-zero vectors, then projection of \( \vec{b} \) on \( \vec{a} \) is \( \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|} \) — Correct. This is the standard result \( |\vec{b}|\cos\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|} \).
(C) The unit vector normal to both vectors \( \vec{a}=\hat{\imath}-\hat{\jmath}-\hat{k} \) and \( \vec{b}=\hat{\imath}+\hat{\jmath}+\hat{k} \) is \( \frac{1}{\sqrt{2}}(-\hat{\jmath}+\hat{k}) \) — Correct. \( \vec{a}\times\vec{b}=-2\hat{\jmath}+2\hat{k} \), which divided by its length \( 2\sqrt{2} \) gives this vector.
(D) The area of triangle formed by adjacent sides represented by vectors \( \vec{a}=3\hat{\imath}+4\hat{\jmath} \) and \( \vec{b}=-5\hat{\imath}+7\hat{\jmath} \) is 41 sq. units — Incorrect. The area is \( \frac{41}{2} \) sq. units; 41 is the area of the parallelogram.