Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | Area of triangle determined by vectors \( (\hat{\imath}+\hat{\jmath}) \) and \( (\hat{\imath}-2\hat{\jmath}+\hat{k}) \) is | I. | 0 |
| B. | The value of \( (\hat{\imath}\times\hat{\jmath}).\hat{\jmath}+(\hat{\jmath}\times\hat{\imath}).\hat{k}+(\hat{k}\times\hat{\imath}).\hat{\jmath} \) is | II. | \( \displaystyle\frac{\sqrt{11}}{2} \) |
| C. | If \( (2\lambda-1)\hat{\imath}+3\hat{\jmath}-\hat{k} \) and \( \hat{\imath}+\lambda\hat{\jmath}+2\hat{k} \) are perpendicular to each other then λ is | III. | 4 |
| D. | Projection of \( 5\hat{\imath}+\hat{\jmath}+4\hat{k} \) on \( 2\hat{\imath}+6\hat{\jmath}+3\hat{k} \) is | IV. | \( \displaystyle\frac{3}{5} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-II, B-I, C-IV, D-III
(A) Area of triangle determined by vectors \( \hat{\imath}+\hat{\jmath} \) and \( \hat{\imath}-2\hat{\jmath}+\hat{k} \)
This matches (II) \( \frac{\sqrt{11}}{2} \).
The area of the triangle with two sides \( \vec{a} \), \( \vec{b} \) is \( \frac{1}{2}|\vec{a}\times\vec{b}| \):
\( \displaystyle \vec{a}\times\vec{b}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\1 & 1 & 0\\1 & -2 & 1\end{vmatrix} \)
\( \displaystyle =(1-0)\hat{\imath}-(1-0)\hat{\jmath}+(-2-1)\hat{k} \)
\( \displaystyle =\hat{\imath}-\hat{\jmath}-3\hat{k} \)
\( \displaystyle |\vec{a}\times\vec{b}|=\sqrt{1+1+9}=\sqrt{11} \)
\( \displaystyle \text{Area}=\frac{\sqrt{11}}{2} \)
(B) The value of \( (\hat{\imath}\times\hat{\jmath})\cdot\hat{\jmath}+(\hat{\jmath}\times\hat{\imath})\cdot\hat{k}+(\hat{k}\times\hat{\imath})\cdot\hat{\jmath} \)
This matches (I) 0.
\( \displaystyle \hat{\imath}\times\hat{\jmath}=\hat{k} \)
\( \displaystyle \hat{\jmath}\times\hat{\imath}=-\hat{k} \)
\( \displaystyle \hat{k}\times\hat{\imath}=\hat{\jmath} \)
\( \displaystyle \hat{k}\cdot\hat{\jmath}+(-\hat{k})\cdot\hat{k}+\hat{\jmath}\cdot\hat{\jmath}=0-1+1=0 \)
(C) λ for perpendicular vectors
This matches (IV) \( \frac{3}{5} \).
Perpendicular vectors have zero dot product:
\( \displaystyle (2\lambda-1)(1)+3\lambda+(-1)(2)=0 \)
\( \displaystyle 5\lambda-3=0 \)
\( \displaystyle \lambda=\frac{3}{5} \)
(D) Projection of \( 5\hat{\imath}+\hat{\jmath}+4\hat{k} \) on \( 2\hat{\imath}+6\hat{\jmath}+3\hat{k} \)
This matches (III) 4. The projection of \( \vec{a} \) on \( \vec{b} \) divides by the length of \( \vec{b} \):
\( \displaystyle \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}=\frac{10+6+12}{\sqrt{4+36+9}} \)
\( \displaystyle =\frac{28}{7}=4 \)
So the matching is A-II, B-I, C-IV, D-III.