JEE Main 2023 — Mathematics Vectors & 3D Geometry

Given,

$$|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|$$∣a+b+c∣=∣a+b−c∣

Now squaring both side we get,

$$\Rightarrow {|\vec{a}+\vec{b}+\vec{c}|}^{2}={|\vec{a}+\vec{b}-\vec{c}|}^{2}$$⇒∣a+b+c∣2=∣a+b−c∣2

$$\Rightarrow {|\vec{a}|}^{2}+{|\vec{b}|}^{2}+{|\vec{c}|}^{2}+2(\vec{a}\cdot \vec{b})+2(\vec{b}\cdot \vec{c})+2(\vec{c}\cdot \vec{a})={|\vec{a}|}^{2}+{|\vec{b}|}^{2}+{|\vec{c}|}^{2}+2(\vec{a}\cdot \vec{b})-2(\vec{b}\cdot \vec{c})-2(\vec{c}\cdot \vec{a})$$⇒∣a∣2+∣b∣2+∣c∣2+2(a⋅b)+2(b⋅c)+2(c⋅a)=∣a∣2+∣b∣2+∣c∣2+2(a⋅b)−2(b⋅c)−2(c⋅a)

$$\Rightarrow 2(\vec{b}\cdot \vec{c})+2(\vec{c}\cdot \vec{a})=-2(\vec{b}\cdot \vec{c})+2(\vec{c}\cdot \vec{a})$$⇒2(b⋅c)+2(c⋅a)=−2(b⋅c)+2(c⋅a)

$$\Rightarrow 4[(\vec{b}\cdot \vec{c})+(\vec{c}\cdot \vec{a})]=0$$⇒4[(b⋅c)+(c⋅a)]=0

Now here given $\vec{b}\cdot \vec{c}=0$b⋅c=0, so

$\vec{c}\cdot \vec{a}=0$c⋅a=0 which means \vec{c}&\vec{a} are perpendicular vector, so statement $(B)$(B) is false,

Now statement$(A):|\vec{a}+\lambda \vec{c}|\geq |\vec{a}|$(A):∣a+λc∣≥∣a∣

$$\Rightarrow {|\vec{a}+\lambda \vec{c}|}^{2}\geq {|\vec{a}|}^{2}$$⇒∣a+λc∣2≥∣a∣2

$$\Rightarrow {|\vec{a}|}^{2}+\lambda {|\vec{c}|}^{2}+2(\vec{a}\cdot \vec{c})\geq {|\vec{a}|}^{2}$$⇒∣a∣2+λ∣c∣2+2(a⋅c)≥∣a∣2

$$\Rightarrow \lambda {|\vec{c}|}^{2}+2(\vec{a}\cdot \vec{c})\geq 0$$⇒λ∣c∣2+2(a⋅c)≥0

$$\Rightarrow \lambda {|\vec{c}|}^{2}\geq 0$$⇒λ∣c∣2≥0

This is not always true, since $\lambda \in R$λ∈R.

Hence, we can say that statement$(A)$(A) is true and statement $(B)$(B) is false.