(C) Let $\vec{a} = \hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b} = \lambda\hat{i} - 4\hat{j} + \hat{k}$.Since the vectors $\vec{a}$ and $\vec{b}$ are or

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(C) Let $\vec{a} = \hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b} = \lambda\hat{i} - 4\hat{j} + \hat{k}$.Since the vectors $\vec{a}$ and $\vec{b}$ are orthogonal, their dot product must be zero, i.e.,$\vec{a} \cdot \vec{b} = 0$.Calculating the dot product:$(\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (\lambda\hat{i} - 4\hat{j} + \hat{k}) = 0$$(1)(\lambda) + (3)(-4) + (4)(1) = 0$$\lambda - 12 + 4 = 0$$\lambda - 8 = 0$$\lambda = 8$.

Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

If the vectors $\hat{i}+3 \hat{j}+4 \hat{k}$ and $\lambda \hat{i}-4 \hat{j}+\hat{k}$ are orthogonal to each other, then $\lambda$ is equal to:

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$5$
B
$-5$
C
$8$
D
$-8$

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Column $I$	Column $II$
$(a)$ Proliferative Phase	$(i)$ Breakdown of endometrium
$(b)$ Secretory Phase	$(ii)$ Follicular phase
$(c)$ Menstruation	$(iii)$ Luteal Phase

If the vectors $\hat{i}+3 \hat{j}+4 \hat{k}$ and $\lambda \hat{i}-4 \hat{j}+\hat{k}$ are orthogonal to each other, then $\lambda$ is equal to:

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Match the items given in Column $I$ with those in Column $II$ and select the correct option given below:

Column $I$ Column $II$

$(a)$ Proliferative Phase $(i)$ Breakdown of endometrium

$(b)$ Secretory Phase $(ii)$ Follicular phase

$(c)$ Menstruation $(iii)$ Luteal Phase