A vector $\vec{a}$ makes equal acute angles on the coordinate axis.Then the projection of vector $\vec{b}=5\hat{i}+7\hat{j}-\hat{k}$ on $\vec{a}$ is

Question

A vector $\vec{a}$  makes equal acute angles on the coordinate axis.Then the projection of vector  $\vec{b}=5\hat{i}+7\hat{j}-\hat{k}$  on  $\vec{a}$  is

cdquestions Admin · Accepted Answer

Since vector $ \vec{a} $ makes equal acute angles with the coordinate axes, we can conclude that its direction ratios are all equal. Thus, we assume that: $$ \vec{a} = \hat{i} + \hat{j} + \hat{k} $$ The projection of vector $ \vec{b} $ on $ \vec{a} $ is given by the formula: $$ \text{Projection of } \vec{b} \text{ on } \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} $$ First, we calculate the dot product $ \vec{a} \cdot \vec{b} $: $$ \vec{a} \cdot \vec{b} = (1)(5) + (1)(7) + (1)(-1) = 5 + 7 - 1 = 11 $$ Next, we calculate the magnitude of $ \vec{a} $: $$ |\vec{a}| = \sqrt{(1^2 + 1^2 + 1^2)} = \sqrt{3} $$ Now, we can find the projection: $$ \text{Projection of } \vec{b} \text{ on } \vec{a} = \frac{11}{\sqrt{3}} $$ Thus, the projection of $ \vec{b} $ on $ \vec{a} $ is $ \frac{11}{\sqrt{3}} $.

Answer

$\frac{11}{15}$

Answer

$\frac{11}{\sqrt3}$

Answer

$\frac{4}{5}$

Answer

$\frac{3}{5\sqrt3}$

List-I	List-II
(A) 4î − 2ĵ − 4k̂	(I) A vector perpendicular to both î + 2ĵ + k̂ and 2î + 2ĵ + 3k̂
(B) 4î − 4ĵ + 2k̂	(II) Direction ratios are −2, 1, 2
(C) 2î − 4ĵ + 4k̂	(III) Angle with the vector î − 2ĵ − k̂ is cos⁻¹(1/√6)
(D) 4î − ĵ − 2k̂	(IV) Dot product with −2î + ĵ + 3k̂ is 10

A vector \(\vec{a}\) makes equal acute angles on the coordinate axis.Then the projection of vector \(\vec{b}=5\hat{i}+7\hat{j}-\hat{k}\) on \(\vec{a}\) is

The Correct Option is B

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