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JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

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The distance between the planes given by r.(i+2j−2k)+5=0\mathbf{r} .(\mathbf{i}+2 \mathbf{j}-2 \mathbf{k})+5=0 and r−(i+2j−2k)−8=0\mathbf{r}-(\mathbf{i}+2 \mathbf{j}-2 \mathbf{k})-8=0 is
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Correct option: B
133\frac{13}{3} unit
Option B is correct because the distance between two parallel planes is calculated using the formula involving the constants and the magnitude of the normal vector.

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A vector which makes equal angles with the vectors 13(i^−2j^+2k^),15(−4i^−3k^)\frac{1}{3}\left( {\hat i - 2\hat j + 2\hat k} \right),\frac{1}{5}\left( { - 4\hat i - 3\hat k} \right), and j^{\hat j}, is If vector a=2i^−3j^+6k^\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}} and vector b=−2i^+2j^−k^\mathbf{b}=-2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, then  Projection of vector a on vector b Projection of vector b on vector a\frac{\text { Projection of vector } \mathbf{a} \text { on vector } \mathbf{b}}{\text { Projection of vector } \mathbf{b} \text { on vector } \mathbf{a}} is equal to The angle between a⃗{\mathbf{\vec a}} and b⃗{\mathbf{\vec b}} is 5π6\frac{{5\pi }}{6} and the projection of a⃗{\mathbf{\vec a}} in the direction of b⃗{\mathbf{\vec b}} is −63\frac{{ - 6}}{{\sqrt 3 }} then ∣a⃗∣\left| {\mathbf{\vec a}} \right| is equal to A velocity 14 m/s\frac{1}{4} \mathrm{~m} / \mathrm{s} is resolved into two components along OAO A and OBO B making angles 30∘30^{\circ} and 45∘45^{\circ} respectively with the given velocity. Then the component along OBO B is Find the equation of the perpendicular drown from the origin to the plane 2x+4y−5z=102x + 4y - 5z = 10 A plane meets the co-ordinate axes at P,QP, Q and RR such that the position vector of the oentroid of △PQR\triangle P Q R is 2i^−5j^+k^2 \hat{i}-5 \hat{j}+\hat{k}. Then the equation of the plane is
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