Which of the following statements is/are TRUE?
A. The sine of the angle between the vectors \( \vec{a}=\hat{\imath}+3\hat{\jmath}-2\hat{k} \) and \( \vec{b}=-2\hat{\imath}+2\hat{\jmath}-4\hat{k} \) is \( \displaystyle\frac{2}{\sqrt{7}} \)
B. The area of the parallelogram PQRS is \( \displaystyle\frac{1}{2}|\overrightarrow{PR}\times\overrightarrow{QS}| \)
C. If \( \vec{a}=5\hat{\imath}-4\hat{\jmath}+3\hat{k} \) then \( |\vec{a}\times\hat{\imath}|^2+|\vec{a}\times\hat{\jmath}|^2+|\vec{a}\times\hat{k}|^2=100 \)
D. Vectors perpendicular to the vectors \( \vec{a}=2\hat{\imath}-\hat{\jmath}+2\hat{k} \) and \( \vec{b}=-\hat{\imath}+4\hat{\jmath}-3\hat{k} \) are given by \( \lambda(-\hat{\imath}+2\hat{\jmath}+2\hat{k}) \) where \( \lambda \) is a scalar.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A, B and C only
For A, with \( \vec{a}=\hat{\imath}+3\hat{\jmath}-2\hat{k} \) and \( \vec{b}=-2\hat{\imath}+2\hat{\jmath}-4\hat{k} \):
\( \displaystyle \vec{a}\cdot\vec{b}=-2+6+8=12 \)
\( \displaystyle |\vec{a}|=\sqrt{14} \)
\( \displaystyle |\vec{b}|=\sqrt{24} \)
\( \displaystyle \cos\theta=\frac{12}{\sqrt{14}\sqrt{24}}=\frac{12}{4\sqrt{21}}=\frac{3}{\sqrt{21}} \)
\( \displaystyle \sin\theta=\sqrt{1-\frac{9}{21}} \)
\( \displaystyle =\sqrt{\frac{12}{21}}=\sqrt{\frac{4}{7}}=\frac{2}{\sqrt{7}} \)
For C, \( \vec{a}\times\hat{\imath}=3\hat{\jmath}+4\hat{k} \), so \( |\vec{a}\times\hat{\imath}|^2=a_2^2+a_3^2 \), and similarly for \( \hat{\jmath} \) and \( \hat{k} \):
\( \displaystyle (a_2^2+a_3^2)+(a_1^2+a_3^2)+(a_1^2+a_2^2)=2|\vec{a}|^2 \)
\( \displaystyle =2(25+16+9)=100 \)
For D, the vectors perpendicular to both are the multiples of the cross product:
\( \displaystyle \vec{a}\times\vec{b}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\2 & -1 & 2\\-1 & 4 & -3\end{vmatrix} \)
\( \displaystyle =(3-8)\hat{\imath}-(-6+2)\hat{\jmath}+(8-1)\hat{k} \)
\( \displaystyle =-5\hat{\imath}+4\hat{\jmath}+7\hat{k} \)
So \( \sin\theta=\frac{2}{\sqrt{7}} \) in A, the sum in C is \( 2|\vec{a}|^2=100 \), and the vectors perpendicular to both vectors in D are the multiples of \( -5\hat{\imath}+4\hat{\jmath}+7\hat{k} \).
Checking the statements
(A) The sine of the angle between the vectors \( \vec{a}=\hat{\imath}+3\hat{\jmath}-2\hat{k} \) and \( \vec{b}=-2\hat{\imath}+2\hat{\jmath}-4\hat{k} \) is \( \frac{2}{\sqrt{7}} \) — Correct. \( \cos\theta=\frac{3}{\sqrt{21}} \) gives \( \sin\theta=\frac{2}{\sqrt{7}} \).
(B) The area of the parallelogram PQRS is \( \frac{1}{2}|\overrightarrow{PR}\times\overrightarrow{QS}| \) — Correct. PR and QS are the diagonals, and the area of a parallelogram is half the magnitude of the cross product of its diagonals.
(C) If \( \vec{a}=5\hat{\imath}-4\hat{\jmath}+3\hat{k} \) then \( |\vec{a}\times\hat{\imath}|^2+|\vec{a}\times\hat{\jmath}|^2+|\vec{a}\times\hat{k}|^2=100 \) — Correct. The sum is \( 2|\vec{a}|^2=2\times 50=100 \).
(D) Vectors perpendicular to the vectors \( \vec{a}=2\hat{\imath}-\hat{\jmath}+2\hat{k} \) and \( \vec{b}=-\hat{\imath}+4\hat{\jmath}-3\hat{k} \) are given by \( \lambda(-\hat{\imath}+2\hat{\jmath}+2\hat{k}) \) where \( \lambda \) is a scalar — Incorrect. \( -\hat{\imath}+2\hat{\jmath}+2\hat{k} \) is not a multiple of \( -5\hat{\imath}+4\hat{\jmath}+7\hat{k} \); its dot product with \( \vec{b} \) is 1 + 8 − 6 = 3, not 0.
Tip: Keep the two area formulas apart: \( |\vec{a}\times\vec{b}| \) with adjacent sides, but \( \frac{1}{2}|\vec{d_1}\times\vec{d_2}| \) with diagonals.