Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | If \( |\vec{a}|=\sqrt{26},\ |\vec{b}|=7 \) and \( |\vec{a}\times\vec{b}|=35 \), then \( |\vec{a}\cdot\vec{b}| \) is equal to | I. | 4 |
| B. | If \( 2\hat{\imath}-3\hat{\jmath}+4\hat{k} \) and \( a\hat{\imath}-6\hat{\jmath}+8\hat{k} \) are collinear, then '\( a \)' is equal to | II. | 2 |
| C. | If \( 2\hat{\imath}+\hat{\jmath}+\hat{k} \) and \( 2\hat{\imath}-4\hat{\jmath}+\lambda\hat{k} \) are perpendicular, then \( \lambda \) is equal to | III. | 7 |
| D. | \( (\hat{\imath}\times\hat{\jmath})\cdot\hat{k}+(\hat{\jmath}\times\hat{k})\cdot\hat{\imath} \) is equal to | IV. | 0 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-III, B-I, C-IV, D-II
(A) \( |\vec{a}|=\sqrt{26},\ |\vec{b}|=7,\ |\vec{a}\times\vec{b}|=35 \)
This matches (III) 7.
By Lagrange's identity \( |\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2=|\vec{a}|^2|\vec{b}|^2 \):
\( \displaystyle 35^2+(\vec{a}\cdot\vec{b})^2=26\times 49 \)
\( \displaystyle (\vec{a}\cdot\vec{b})^2=1274-1225=49 \)
\( \displaystyle |\vec{a}\cdot\vec{b}|=7 \)
(B) \( 2\hat{\imath}-3\hat{\jmath}+4\hat{k} \) and \( a\hat{\imath}-6\hat{\jmath}+8\hat{k} \) collinear
This matches (I) 4.
Collinear vectors have proportional components:
\( \displaystyle \frac{a}{2}=\frac{-6}{-3}=\frac{8}{4}=2 \)
\( \displaystyle a=4 \)
(C) \( 2\hat{\imath}+\hat{\jmath}+\hat{k} \) and \( 2\hat{\imath}-4\hat{\jmath}+\lambda\hat{k} \) perpendicular
This matches (IV) 0.
\( \displaystyle 2(2)+1(-4)+1(\lambda)=0 \)
\( \displaystyle \lambda=0 \)
(D) \( (\hat{\imath}\times\hat{\jmath})\cdot\hat{k}+(\hat{\jmath}\times\hat{k})\cdot\hat{\imath} \)
This matches (II) 2.
\( \displaystyle \hat{\imath}\times\hat{\jmath}=\hat{k},\ \hat{\jmath}\times\hat{k}=\hat{\imath} \)
\( \displaystyle \hat{k}\cdot\hat{k}+\hat{\imath}\cdot\hat{\imath}=1+1=2 \)
So the matching is A-III, B-I, C-IV, D-II.