Verified Solution • 2025–26 Session moderate

QJEEM23Ma5b74: JEE Main 2023 Physics - Class 11 JEEM23Ma5b74 Solution: Let . Then is equal to :

Let a→=i^+2j^+λk^,b→=3i^−5j^−λk^,a→ . c→=7,2b→ . c→+43=0,a→×c→=b→×c→\overrightarrow a = \widehat i + 2\widehat j + \lambda \widehat k,\overrightarrow b = 3\widehat i - 5\widehat j - \lambda \widehat k,\overrightarrow a \,.\,\overrightarrow c = 7,2\overrightarrow b \,.\,\overrightarrow c + 43 = 0,\overrightarrow a \times \overrightarrow c = \overrightarrow b \times \overrightarrow c . Then ∣a→ . b→∣\left| {\overrightarrow a \,.\,\overrightarrow b } \right| is equal to :

Step-by-Step Verified Solution

amp;a⃗=i^+2j^+λk^,b⃗=3i^−5j^−λk^,a⃗⋅c⃗=7amp;a⃗×c⃗−b⃗×c⃗=0→amp;(a⃗−b⃗)×c⃗=0⇒(a⃗−b⃗) is paralleled to c⃗amp;a⃗−b⃗=μc⃗, where μ is a scalar amp;−2i^+7j^+2λk^=μ⋅c→\begin{aligned}& \vec{a}=\hat{i}+2 \hat{j}+\lambda \hat{k}, \vec{b}=3 \hat{i}-5 \hat{j}-\lambda \hat{k}, \vec{a} \cdot \vec{c}=7 \\\\& \vec{a} \times \vec{c}-\vec{b} \times \vec{c}=\overrightarrow{0} \\\\& (\vec{a}-\vec{b}) \times \vec{c}=0 \Rightarrow(\vec{a}-\vec{b}) \text { is paralleled to } \vec{c} \\\\& \vec{a}-\vec{b}=\mu \vec{c}, \text { where } \mu \text { is a scalar } \\\\& -2 \hat{i}+7 \hat{\mathrm{j}}+2 \lambda \hat{k}=\mu \cdot \overrightarrow{\mathrm{c}}\end{aligned}

Now a⃗⋅c→=7\vec{a} \cdot \overrightarrow{\mathbf{c}}=7 gives 2λ2+12=7μ2 \lambda^2+12=7 \mu

And b⃗⋅c⃗=−432\vec{b} \cdot \vec{c}=-\frac{43}{2} gives 4λ2+82=43μ4 \lambda^2+82=43 \mu

μ=2\mu=2 and λ2=1\lambda^2=1

∣a→⋅b→∣=8|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|=8

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