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If a⃗,b⃗\vec a, \vec b and c⃗\vec c are vectors such that ∣a⃗∣=5,∣b⃗∣=12,∣c⃗∣=13|\vec a|=5, |\vec b|=12, |\vec c|=13 and a⃗+b⃗+c⃗=0⃗\vec a+\vec b+\vec c=\vec 0, then

Match List-I with List-II

LIST-ILIST-II
A. a⃗.c⃗+b⃗.c⃗\vec{a}.\vec{c} + \vec{b}.\vec{c}I. −25-25
B. a⃗.b⃗+a⃗.c⃗\vec{a}.\vec{b} + \vec{a}.\vec{c}II. −338-338
C. a⃗.b⃗+b⃗.c⃗\vec{a}.\vec{b} + \vec{b}.\vec{c}III. −169-169
D. 2(a⃗.b⃗+b⃗.c⃗+c⃗.a⃗)2\left(\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a}\right)IV. −144-144

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 4

Since the sum is zero, a⃗+b⃗=−c⃗\vec a+\vec b=-\vec c, so A gives (a⃗+b⃗)⋅c⃗=−∣c⃗∣2=−169(\vec a+\vec b)\cdot\vec c=-|\vec c|^2=-169 (III).

Similarly B gives −∣a⃗∣2=−25-|\vec a|^2=-25 (I)

C gives −∣b⃗∣2=−144-|\vec b|^2=-144 (IV).

Squaring the zero sum, D equals −(25+144+169)=−338-(25+144+169)=-338 (II).

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