Match List I with List II
| LIST I | LIST II | ||
| A. | \(|\vec{a}\times\hat{\imath}|^2+|\vec{a}\times\hat{\jmath}|^2+|\vec{a}\times\hat{k}|^2=\) | I. | \(|\vec{a}|^2\) |
| B. | \(\displaystyle\frac{(\vec{a}\cdot\vec{b})^2+|\vec{a}\times\vec{b}|^2}{|\vec{b}|^2}=\) | II. | \(\sqrt{3}|\vec{a}|\) |
| C. | \(\vec{a}, \vec{b}\) and \(\vec{c}\) are three mutually perpendicular vectors of equal magnitude then \(|\vec{a}+\vec{b}+\vec{c}|=\) | III. | \(\displaystyle\frac{1}{2}|\vec{a}\times\vec{b}|\) |
| D. | Area of parallelogram with diagonals \(\vec{a}\) and \(\vec{b}\) | IV. | \(2|\vec{a}|^2\) |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-IV, B-I, C-II, D-III
The key result is Lagrange’s identity \(|\vec{u}\times\vec{v}|^2=|\vec{u}|^2|\vec{v}|^2-(\vec{u}\cdot\vec{v})^2\).
(A) \(|\vec{a}\times\hat{\imath}|^2+|\vec{a}\times\hat{\jmath}|^2+|\vec{a}\times\hat{k}|^2\)
This matches (IV) \(2|\vec{a}|^2\). With \(\vec{a}=a_1\hat{\imath}+a_2\hat{\jmath}+a_3\hat{k}\), Lagrange’s identity gives \(|\vec{a}\times\hat{\imath}|^2=|\vec{a}|^2-a_1^2\), and similarly for the others:
\(3|\vec{a}|^2-(a_1^2+a_2^2+a_3^2)=3|\vec{a}|^2-|\vec{a}|^2=2|\vec{a}|^2\)
(B) \(\frac{(\vec{a}\cdot\vec{b})^2+|\vec{a}\times\vec{b}|^2}{|\vec{b}|^2}\)
This matches (I) \(|\vec{a}|^2\), since by Lagrange’s identity the numerator is \(|\vec{a}|^2|\vec{b}|^2\).
(C) \(|\vec{a}+\vec{b}+\vec{c}|\) for mutually perpendicular vectors of equal magnitude
This matches (II) \(\sqrt{3}|\vec{a}|\). All dot products are zero, so
\(|\vec{a}+\vec{b}+\vec{c}|^2=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2=3|\vec{a}|^2\)
(D) Area of the parallelogram with diagonals \(\vec{a}\) and \(\vec{b}\)
This matches (III) \(\frac{1}{2}|\vec{a}\times\vec{b}|\). If the sides are \(\vec{p}\) and \(\vec{q}\), then \(\vec{a}=\vec{p}+\vec{q}\), \(\vec{b}=\vec{q}-\vec{p}\), and
\(\vec{a}\times\vec{b}=(\vec{p}+\vec{q})\times(\vec{q}-\vec{p})=\vec{p}\times\vec{q}-\vec{q}\times\vec{p}=2\,\vec{p}\times\vec{q}\)
So the area \(|\vec{p}\times\vec{q}|=\frac{1}{2}|\vec{a}\times\vec{b}|\).